Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Consider the following nonlinear, autonomous vector field on \( {\mathbb{R}}^{2} \) :\n\n\[ \dot{x} = \mu - {x}^{2} \]\n\n\[ \dot{y} = - y,\;\left( {x, y}\right) \in {\mathbb{R}}^{2} \]\n\n(8.1)\n\nwhere \( \mu \) is a (real) parameter. The equilibrium points of (8.1) are given by:\n\n\[ \left( {x, y}\right) = \left( {... | The Jacobian of the vector field evaluated at each equilibrium point is given by:\n\n\[ \left( {\sqrt{}\bar{\mu },0}\right) : \;\left( \begin{matrix} - 2\sqrt{}\bar{\mu } & 0 \\ 0 & - 1 \end{matrix}\right) ,\n\n(8.3) \]\n\nfrom which it follows that the equilibria are hyperbolic and asymptotically stable for \( \mu > 0... | Yes |
Consider the following nonlinear, autonomous vector field on \( {\mathbb{R}}^{2} \) :\n\n\[ \dot{x} = {\mu x} - {x}^{2} \]\n\n\[ \dot{y} = - y,\;\left( {x, y}\right) \in {\mathbb{R}}^{2}, \] | The equilibrium points of (8.5) are given by:\n\n\[ \left( {x, y}\right) = \left( {0,0}\right) ,\left( {\mu ,0}\right) . \]\n\nThe Jacobian of the vector field evaluated at each equilibrium point is given by:\n\n\[ \left( {0,0}\right) \;\left( \begin{matrix} \mu & 0 \\ 0 & - 1 \end{matrix}\right) \]\n\n\[ \left( {\mu ,... | Yes |
Consider the following nonlinear, autonomous vector field on \( {\mathbb{R}}^{2} \) :\n\n\[ \dot{x} = {\mu x} - {x}^{3} \]\n\n\[ \dot{y} = - y,\;\left( {x, y}\right) \in {\mathbb{R}}^{2}, \]\n\n(8.9)\n\nwhere \( \mu \) is a (real) parameter. The equilibrium points of (8.9) are given by:\n\n\[ \left( {x, y}\right) = \le... | The Jacobian of the vector field evaluated at each equilibrium point is given by:\n\n\[ \left( \begin{matrix} \mu & 0 \\ 0 & - 1 \end{matrix}\right) \]\n\n(8.11)\n\n\[ \left( {\pm \sqrt{\mu },0}\right) \;\left( \begin{matrix} - {2\mu } & 0 \\ 0 & - 1 \end{matrix}\right) \]\n\n(8.12)\n\nfrom which it follows that \( \le... | Yes |
Consider the following nonlinear, autonomous vector field on \( {\mathbb{R}}^{2} \) :\n\n\[ \n\dot{x} = {\mu x} + {x}^{3} \n\]\n\n\[ \n\dot{y} = - y,\;\left( {x, y}\right) \in {\mathbb{R}}^{2}, \n\]\n\n(8.13)\n\nwhere \( \mu \) is a (real) parameter. The equilibrium points of (8.9) are given by:\n\n\[ \n\left( {x, y}\r... | Figure 8.6: Bifurcation diagram for (8.9) in the \( x - y \) plane for \( \mu < 0,\mu = 0 \), and\n\n\( \mu < 0 \), and do not exist for \( \mu > 0 \) . These two curves of fixed points pass through zero at \( \mu = 0 \), at which there is only one, nonhyperbolic fixed point.\n\nIn Fig. 8.7 we show the bifurcation diag... | No |
We consider the one dimensional autonomous vector field:\n\n\[ \n\\dot{x} = \\mu - {x}^{3},\\;x \\in \\mathbb{R} \n\]\n\n(8.17)\n\nwhere \( \\mu \) is a parameter. This vector field has a nonhyperbolic fixed point at \( x = 0 \) for \( \\mu = 0 \) . The curve of fixed points in the \( \\mu - x \) plane is given by \( \... | In fig. 8.9 we plot the fixed points as a function of \( \\mu \) .\n\n\n\nFigure 8.9: Bifurcation diagram for \( \\left( {8.17}\\right) \) .\n\nWe see that there is no change in the number or stability of the fixed poi... | Yes |
Consider the following one dimensional autonomous vector field depending on a parameter \( \mu \) :\n\n\[ \dot{x} = \mu - \frac{{x}^{2}}{2} + \frac{{x}^{3}}{3},\;x \in \mathbb{R}. \] | The fixed points of this vector field are given by:\n\n\[ \mu = \frac{{x}^{2}}{2} - \frac{{x}^{3}}{3} \] | No |
Example 28. Consider the following one dimensional autonomous vector field depending on a parameter \( \mu \) :\n\n\[ \n\dot{x} = {\mu x} - \frac{{x}^{3}}{2} + \frac{{x}^{4}}{3} \n\]\n\n\[ \n= x\left( {\mu - \frac{{x}^{2}}{2} + \frac{{x}^{3}}{3}}\right) \n\]\n\n(9.19)\n\nThe fixed points of this vector field are given ... | and are plotted in the \( \mu - x \) plane in Fig. 9.6.\n\n\n\nFigure 9.6: Fixed points of (9.19) plotted in the \( \mu - x \) plane.\n\nIn this example we see that there is a pitchfork bifurcation and a saddle-node bi... | Yes |
Consider the following linear, autonomous vector field on \( {\mathbb{R}}^{c} \times \) \( {\mathbb{R}}^{s} \): \n\n\[ \n\dot{x} = {Ax} \n\] \n\n\[ \n\dot{y} = {By},\;\left( {x, y}\right) \in {\mathbb{R}}^{c} \times {\mathbb{R}}^{s}, \n\] \n\n(10.1) \n\nwhere \( A \) is a \( c \times c \) matrix of real numbers having ... | This follows from the nature of the eigenvalues of \( B \), and the properties that \( x \) and \( y \) are decoupled in (10) and that it is linear. More precisely, the solution of (10) is given by: \n\n\[ \n\left( \begin{array}{l} x\left( {t,{x}_{0}}\right) \\ y\left( {t,{y}_{0}}\right) \end{array}\right) = \left( \be... | Yes |
Theorem 5 (Existence and Restricted Dynamics). There exists a \( {C}^{r} \) center manifold of \( \left( {x, y}\right) = \left( {0,0}\right) \) for (10.4). The dynamics of (10.4) restricted to the center manifold is given by: | \[ \dot{u} = {Au} + f\left( {u, h\left( u\right) }\right) ,\;u \in {\mathbb{R}}^{c}, \] \( \left( {10.10}\right) \) ## for \( \left| u\right| \) sufficiently small. | Yes |
We consider the following autonomous vector field on the plane:\n\n\[ \n\\dot{x} = {x}^{2}y - {x}^{5} \n\]\n\n\[ \n\\dot{y} = - y + {x}^{2},\;\\left( {x, y}\\right) \\in {\\mathbb{R}}^{2}. \n\]\n\n\\( \\left( {10.18}\\right) \\)\n\nor, in matrix form:\n\n\[ \n\\left( \\begin{array}{l} \\dot{x} \\\\ \\dot{y} \\end{array... | The Jacobian associated with the linearization about this fixed point is:\n\n\[ \n\\left( \\begin{matrix} 0 & 0 \\\\ 0 & - 1 \\end{matrix}\\right) \n\]\n\nwhich is nonhyperbolic, and therefore the linearization does not suffice to determine stability.\n\nThe vector field is in the form of (10.4)\n\n\[ \n\\dot{x} = {Ax}... | Yes |
Example 31. We consider the following autonomous vector field on the plane:\n\n\[ \n\\dot{x} = {xy} \n\]\n\n\[ \n\\dot{y} = - y + {x}^{3},;\\left( {x, y}\\right) \\in {\\mathbb{R}}^{2}, \n\]\n\n\\( \\left( {10.27}\\right) \\)\n\nor, in matrix form:\n\n\[ \n\\left( \\begin{array}{l} \\dot{x} \\\\ \\dot{y} \\end{array}\\... | The vector field is in the form of (10)\n\n\[ \n\\dot{x} = {Ax} + f\\left( {x, y}\\right) , \n\]\n\n\[ \n\\dot{y} = {By} + g\\left( {x, y}\\right) ,;\\left( {x, y}\\right) \\in \\mathbb{R} \\times \\mathbb{R}, \n\]\n\n(10.29)\n\nwhere\n\n\[ \nA = 0, B = - 1, f\\left( {x, y}\\right) = {xy}, g\\left( {x, y}\\right) = {x}... | No |
Consider the one dimensional, autonomous linear vector field:\n\n\\[ \n\\dot{x} = {ax},\\;x, a \\in \\mathbb{R}. \n\\] | We often solve problems in mathematics by transforming them into simpler problems that we already know how to solve. Towards this end, we introduce the following (time-dependent) transformation of variables:\n\n\\[ \nx = u{e}^{at}. \n\\]\n\nDifferentiating this expression with respect to \\( t \\), and using (B.1), giv... | Yes |
Consider the following linear inhomogeneous nonautonomous ODE (due to the presence of the term \( b\left( t\right) \) ):\n\n\[ \n\dot{x} = {ax} + b\left( t\right) ,\;a, x \in \mathbb{R}, \n\] | We will use exactly the same strategy and change of coordinates as in the previous example:\n\n\[ \nx = u{e}^{at}. \n\]\n\nDifferentiating this expression with respect to \( t \), and using (B.8), gives:\n\n\[ \n\dot{u} = {e}^{-{at}}b\left( t\right) \n\]\n\nIntegrating (B.10) gives:\n\n\[ \nu\left( t\right) = u\left( 0... | Yes |
Consider the one dimensional, nonautonomous linear vector field:\n\n\[ \n\\dot{x} = a\\left( t\\right) x,\;x \\in \\mathbb{R}, \n\]\n\nwhere \( a\\left( t\\right) \) is a scalar valued function of \( t \) whose precise properties will be considered later. | We introduce the following (time-dependent) transformation of variables (compare with (36)):\n\n\[ \nx = u{e}^{{\\int }_{0}^{t}a\\left( {t}^{\\prime }\\right) d{t}^{\\prime }}. \n\]\n\nDifferentiating this expression with respect to \( t \), and substituting (B.13) into the result gives:\n\n\[ \n\\dot{x} = \\dot{u}{e}^... | Yes |
Consider the one dimensional inhomogeneous nonautonomous linear vector field:\n\n\\[ \n\\dot{x} = a\\left( t\\right) x + b\\left( t\\right) ,\\;x \\in \\mathbb{R}, \n\\]\n\nwhere \\( a\\left( t\\right), b\\left( t\\right) \\) are scalar valued functions whose required properties will be considered at the end of this ex... | We make the same transformation as (B.14):\n\n\\[ \nx = u{e}^{{\\int }_{0}^{t}a\\left( {t}^{\\prime }\\right) d{t}^{\\prime }} \n\\]\n\nfrom which we obtain:\n\n\\[ \n\\dot{u} = b\\left( t\\right) {e}^{-{\\int }_{0}^{t}a\\left( {t}^{\\prime }\\right) d{t}^{\\prime }}. \n\\]\n\nIntegrating this expression gives:\n\n\\[ ... | Yes |
Consider the \( n \) dimensional autonomous linear vector field:\n\n\[ \dot{x} = {Ax},\;x \in {\mathbb{R}}^{n}, \] | where \( A \) is a \( n \times n \) matrix of real numbers. We make the following transformation of variables (compare with ):\n\n\[ x = {e}^{At}u \]\n\nDifferentiating this expression with respect to \( t \), and using (B.25), gives:\n\n\[ \dot{u} = 0 \]\n\nIntegrating this expression gives:\n\n\[ u\left( t\right) = u... | Yes |
Example 37. Consider the \( n \) dimensional inhomogeneous nonautonomous linear vector field:\n\n\[ \dot{x} = {Ax} + g\left( t\right) ,\;x \in {\mathbb{R}}^{n}, \] | We use the same transformation as in the previous example:\n\n\[ x = {e}^{At}u \]\n\nDifferentiating this expression with respect to \( t \), and using (B.31), gives:\n\n\[ \dot{u} = {e}^{-{At}}g\left( t\right) \]\n\nfrom which it follows that:\n\n\[ u\left( t\right) = u\left( 0\right) + {\int }_{0}^{t}{e}^{-A{t}^{\pri... | Yes |
Consider the following autonomous vector field on \( {\mathbb{R}}^{2} \) :\n\n\[ \dot{x} = y \]\n\n\[ \dot{y} = - x - {\delta y},\;\delta \geq 0,\;\left( {x, y}\right) \in {\mathbb{R}}^{2}. \]\n\n(C.5) | For \( \delta = 0\left( {\mathrm{C} \cdot 5}\right) \) has the form of (C.1):\n\n\[ \dot{x} = y \]\n\n\[ \dot{y} = - x,\;\left( {x, y}\right) \in {\mathbb{R}}^{2}. \]\n\n(C.6)\n\nwith\n\n\[ E = \frac{{y}^{2}}{2} + \frac{{x}^{2}}{2} \]\n\n(C.7)\n\nIt is easy to verify that \( \frac{dE}{dt} = 0 \) along trajectories of (... | Yes |
Consider the following autonomous vector field on \( {\mathbb{R}}^{2} \) :\n\n\[ \n\dot{x} = y \]\n\n\[ \n\dot{y} = x - {x}^{3} - {\delta y},\;\delta \geq 0,\;\left( {x, y}\right) \in {\mathbb{R}}^{2}.\n\]\n\n(C.9)\n\nFor \( \delta = 0 \) (C.9) has the form of (C.1):\n\n\[ \n\dot{x} = y \]\n\n\[ \n\dot{y} = x - {x}^{3}... | The function \( E \) can be used to apply the LaSalle invariance principle to conclude that for \( \delta > 0 \) all trajectories approach one of the three equilibria as \( t \rightarrow \infty \) . | Yes |
Example 40. We now consider an example which was exercise 1b from Problem Set 8.\n\n\[ \dot{x} = {\mu x} + {10}{x}^{2} \]\n\n\[ \dot{\mu } = 0 \]\n\n\[ \dot{y} = x - {2y},\;\left( {x, y}\right) \in {\mathbb{R}}^{2},\mu \in \mathbb{R}. \] | The Jacobian associated with the linearization about \( \left( {x,\mu, y}\right) = \left( {0,0,0}\right) \) is given by:\n\n\[ \left( \begin{array}{rrr} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & - 2 \end{array}\right) \]\n\nIt is easy to check that the eigenvalues of this matrix are 0,0, and \( - 2 \) (as we would have expecte... | Yes |
We consider the Hamiltonian:\n\n\[ H\left( {q, p}\right) = \frac{\lambda }{2}\left( {{p}^{2} - {q}^{2}}\right) = \frac{\lambda }{2}\left( {p - q}\right) \left( {p + q}\right) ,\;\left( {q, p}\right) \in {\mathbb{R}}^{2} \] | From this Hamiltonian, we derive Hamilton’s equations:\n\n\[ \dot{q} = \frac{\partial H}{\partial p}\left( {q, p}\right) = {\lambda p} \]\n\n\[ \dot{p} = - \frac{\partial H}{\partial q}\left( {q, p}\right) = {\lambda q} \]\n\nor in matrix form:\n\n\[ \left( \begin{array}{l} \dot{q} \\ \dot{p} \end{array}\right) = \left... | Yes |
We consider the Hamiltonian:\n\n\[ H\left( {q, p}\right) = \frac{\omega }{2}\left( {{p}^{2} + {q}^{2}}\right) ,\;\left( {q, p}\right) \in {\mathbb{R}}^{2} \] | From this Hamiltonian, we derive Hamilton’s equations:\n\n\[ \dot{q} = \frac{\partial H}{\partial p}\left( {q, p}\right) = {\omega p} \]\n\n\[ \dot{p} = - \frac{\partial H}{\partial q}\left( {q, p}\right) = - {\omega q} \]\n\nor, in matrix form:\n\n\[ \left( \begin{array}{l} \dot{q} \\ \dot{p} \end{array}\right) = \lef... | Yes |
Example 43 (Hamiltonian saddle-node bifurcation). We consider the Hamiltonian:\n\n\[ \nH\left( {q, p}\right) = \frac{{p}^{2}}{2} - {\lambda q} + \frac{{q}^{3}}{3},\;\left( {q, p}\right) \in {\mathbb{R}}^{2}. \n\]\n\n(E.13)\n\nwhere \( \lambda \) is considered to be a parameter that can be varied. From this Hamiltonian,... | Next we examine stability of the fixed points. The Jacobian of (E.14) is given by:\n\n\[ \n\left( \begin{matrix} 0 & 1 \\ - {2q} & 0 \end{matrix}\right) \n\]\n\n(E.16)\n\nThe eigenvalues of this matrix are:\n\n\[ \n{\lambda }_{1,2} = \pm \sqrt{-{2q}} \n\]\n\nHence \( \left( {q, p}\right) = \left( {-\sqrt{\lambda },0}\r... | Yes |
Example 44 (Hamiltonian pitchfork bifurcation). We consider the Hamiltonian:\n\n\[ \nH\left( {q, p}\right) = \frac{{p}^{2}}{2} - \lambda \frac{{q}^{2}}{2} + \frac{{q}^{4}}{4} \]\n\n(E.17)\n\nwhere \( \lambda \) is considered to be a parameter that can be varied. From this Hamiltonian, we derive Hamilton's equations:\n\... | Next we examine stability of the fixed points. The Jacobian of (E.18) is given by:\n\n\[ \n\left( \begin{matrix} 0 & 1 \\ \lambda - 3{q}^{2} & 0 \end{matrix}\right) \]\n\n(E.20)\n\nThe eigenvalues of this matrix are:\n\n\[ \n{\lambda }_{1,2} = \pm \sqrt{\lambda - 3{q}^{2}} \]\n\nHence \( \left( {q, p}\right) = \left( {... | Yes |
Consider the autonomous vector field on the cylinder:\n\n\[ \dot{r} = 0, \]\n\n\[ \dot{\theta } = r,\;\left( {r,\theta }\right) \in {\mathbb{R}}^{ + } \times {S}^{1} \] | The flow generated by this vector field is given by:\n\n\[ {\phi }_{t}\left( {{r}_{0},{\theta }_{0}}\right) = \left( {{r}_{0},{r}_{0}t + {\theta }_{0}}\right) \]\n\nNote that \( r \) is constant in time. This implies that any annulus is an invariant set. In particular, choose any \( {r}_{1} < {r}_{2} \). Then the annul... | Yes |
Consider the following two dimensional autonomous vector field on the cylinder:\n\n\[ \n\dot{r} = \sin \frac{\pi }{r} \n\]\n\n\[ \n\dot{\theta } = r,\;\left( {r,\theta }\right) \in {\mathbb{R}}^{ + } \times {S}^{1}. \n\]\n\nEquilibrium points of the \( \dot{r} \) component of this vector field correspond to periodic or... | This is given by:\n\n\[ \n- \frac{\pi }{{r}^{2}}\cos \frac{\pi }{r} \n\]\n\nand evaluating this on the periodic orbits gives;\n\n\[ \n- \frac{\pi }{{n}^{2}}{\left( -1\right) }^{n} \n\]\n\nTherefore all of these periodic orbits are hyperbolic and stable for \( n \) even and unstable for \( n \) odd. This is an example o... | Yes |
Proposition 3.1.5. Suppose \( p \) is a prime and \( a, b \in \mathbb{Z} \) . If \( p \mid {ab} \) then \( p \mid \) or \( p \mid b \) . | Proof. Notice that \( \gcd \left( {p, a}\right) \mid p \), therefore \( \gcd \left( {p, a}\right) \) is either 1 or \( p \) since \( p \) is prime. But also \( \gcd \left( {a, p}\right) \mid a \), so either \( p \mid a \) or \( \gcd \left( {a, p}\right) = 1 \) . If \( p \mid a \), we are done. If not, since therefore \... | Yes |
Proposition 3.3.2. Given relatively prime numbers \( n \in \mathbb{N} \) and \( a \in \mathbb{Z} \) with \( n \geq 2 \) , \( {\operatorname{ord}}_{n}\left( a\right) \) is well-defined. | Proof. The problem in the definition of \( {\operatorname{ord}}_{n}\left( a\right) \) might be that there might not be any value of \( k \in \mathbb{N} \) at all for which \( {a}^{k} \equiv 1\left( {\;\operatorname{mod}\;n}\right) \) .\n\nBut notice that this is a congruence, so we are really only concerned with the el... | Yes |
Proposition 4.4.6. Let notation be as in the above definition of the RSA cryptosystem. Then for any message \( m \in \mathcal{M},{d}_{\left( n, d\right) }\left( {{e}_{\left( n, e\right) }\left( m\right) }\right) = m \) . | Proof. Any \( m \in \mathcal{M} \) represents a class in \( \mathbb{Z}/n\mathbb{Z} \), and we will blur the distinction between the class and its representative \( m \) satisfying \( 0 \leq m < n \) .\n\nWe have built \( d \) as \( d = {e}^{-1}\left( {{\;\operatorname{mod}\;\phi }\left( n\right) }\right) \) . This mean... | No |
THEOREM 5.1.1. Suppose \( n \in \mathbb{N} \) and \( a \in \mathbb{Z} \) satisfy \( n \geq 2 \) and \( \gcd \left( {a, n}\right) = 1 \) . Then \( {a}^{k} \equiv 1\left( {\;\operatorname{mod}\;n}\right) \) for \( k \in \mathbb{N} \) iff \( {\operatorname{ord}}_{n}\left( a\right) \mid k. \) | Proof. One direction is very easy: if \( k \in \mathbb{N} \) satisfies \( {\operatorname{ord}}_{n}\left( a\right) \mid k \) then \( \exists d \in \mathbb{N} \) such that \( k = {\operatorname{ord}}_{n}\left( a\right) \cdot d \) and thus\n\n\[ \n{a}^{k} = {a}^{{\operatorname{ord}}_{n}\left( a\right) \cdot d} = {\left( {... | Yes |
Proposition 5.5.2. When Alice and Bob follow the DHKE protocol, they both compute the same shared key; i.e., DHKE works. | Proof. There's very little to check here: using the notation of the definition and, in the very middle, the commutativity of multiplication, we have\n\n\[ \n{B}^{\alpha } = {\left( {r}^{\beta }\right) }^{\alpha } = {r}^{\beta \cdot \alpha } = {r}^{\alpha \cdot \beta } = {\left( {r}^{\alpha }\right) }^{\beta } = {A}^{\b... | Yes |
Proposition 5.6.2. With the notation as above in Definition 5.6.1 we have\n\n\[ \n{d}_{\left( p, r,\alpha \right) }\left( {{e}_{\left( p, r, a\right) }\left( m\right) }\right) = m\;\forall m \in \mathcal{M}. \n\] | Proof. Just compute:\n\n\[ \n{d}_{\left( p, r,\alpha \right) }\left( {{e}_{\left( p, r, a\right) }\left( m\right) }\right) \equiv m \cdot {a}^{\beta }\left( {\;\operatorname{mod}\;p}\right) \cdot {\left( {r}^{\beta }\right) }^{p - 1 - \alpha }\;\left( {\;\operatorname{mod}\;p}\right) \n\]\n\n\[ \n\equiv m \cdot {\left(... | Yes |
Proposition 5.6.4. Using the notation as above, Bob will accept all signed messages produced by Alice. | Proof. Assuming the signed message \( \left( {m, x, y}\right) \) was produced by Alice as above, we compute:\n\n\[ \n{a}^{x}{x}^{y} \equiv {a}^{{r}^{\gamma }}{\left( {r}^{\gamma }\right) }^{{\gamma }^{-1}\left( {m - \alpha {r}^{\gamma }}\right) }\;\left( {\;\operatorname{mod}\;p}\right) \n\]\n\n\[ \n\equiv {\left( {r}^... | Yes |
Example 2.1.7 Convert the cake division payoffs so that the payoff vectors sum to zero (rather than 100). | The solution is given in Table 2.1.8.\n\nTable 2.1.8 Zero-sum payoff matrix for Cake Cutting game.\n\n<table><tr><td rowspan=\ | Yes |
Example 2.2.1 A Simpler Payoff Matrix. Consider the zero-sum game with payoff matrix in Table 2.2.2. Note that for simplicity our payoff matrix contains only the payoffs and not the strategy names; but Player 1 still chooses a row and Player 2 still chooses a column. | If we know we are playing a zero-sum game, then the use of ordered pairs seems somewhat redundant: If Player 1 wins 1, then we know that Player 2 must lose 1 (win -1). Thus, if we KNOW we are playing a zero-sum game, we can simplify our notation by just using Player 1's payoffs. The above matrix in Table 2.2.2 can be s... | No |
Determine which row Player 1 should choose. Is there any situation in which Player 1 would choose the other row? | In Example 2.2.6, no matter what Player 2 does, Player 1 would always choose Row 1, since every payoff in Row 1 is greater than or equal to the corresponding payoff in Row 2\n\n( {1 \geq - 1,0 \geq - 2,2 \geq 2} ) . In Example 2.2.8, this is not the case: if Player 2 were to choose Column 3, then Player 1 would prefer ... | Yes |
Example 2.3.1 Drawing a Particular Suit. Given a standard deck of playing cards, what is the probability of drawing a heart? | Answer: You might say since there are four suits, and one of the suits is hearts, you have a probability of \( \frac{1}{4} \) . You’d be correct, but be careful with this reasoning. This works because each suit has the same number of cards, so each suit is equally likely. Another way the calculate the probability is to... | Yes |
Example 2.3.2 A Card is Missing. Now suppose the ace of spades is missing from the deck. What is the probability of drawing a heart? | Answer: As before, there are still four suits in the deck, so it might be tempting to say the probability is still \( \frac{1}{4} \) . But we’d be wrong! Each suit is no longer equally likely since, it is slightly less likely that we draw a spade. Each individual card is still equally likely, though. So now\n\n\[ P\lef... | Yes |
Solution Theorem for Zero-Sum Games. Every equilibrium point of a two-person zero-sum game has the same value. | Let’s start with the \( 2 \times 2 \) case. We will use a proof by contradiction. We will assume the theorem is false and show that we get a logical contradiction. Once we reach a logical contradiction (a statement that is both true and false), we can conclude we were wrong to assume the theorem was false; hence, the t... | No |
Example 3.7.1 Undercut. Each player chooses a number 1-5. If the two numbers don't differ by 1, then each player adds their own number to their score. If the two numbers differ by 1, then the player with the lower number adds both numbers to his or her score; the player with the higher number gets nothing. (From Dougla... | For example, suppose in round one Player 1 chooses 4; Player 2 chooses 4. Each player keeps their own number. The score is now 4-4. In the next round, Player 1 chooses 2, Player 2 chooses 5. The score would now be 6-9. In the third round Player 1 chooses 4, Player 2 chooses 5. Now Player 1 gets both numbers, making the... | No |
Lemma 4 (Legendre). \( {e}_{p}\left( n\right) = \left\lfloor \frac{n}{p}\right\rfloor + \left\lfloor \frac{n}{{p}^{2}}\right\rfloor + \left\lfloor \frac{n}{{p}^{3}}\right\rfloor + \cdots \) . | In fact \( \left\lfloor \frac{n}{p}\right\rfloor \) is the number of multiples of \( p \) in \( n \) !, the term \( \left\lfloor \frac{n}{{p}^{2}}\right\rfloor \) adds the additional contribution of the multiples of \( {p}^{2} \), and so on, e.g., \[ {e}_{3}\left( {30}\right) = \left\lfloor \frac{30}{3}\right\rfloor + ... | Yes |
Theorem 1. \( \\mathop{\\prod }\\limits_{{p \\leq n}}p < {4}^{n} \) . | The proof is by induction on \( n \) . We assume the theorem true for integers \( < n \) and consider the cases \( n = {2m} \) and \( n = {2m} + 1 \) . If \( n = {2m} \) then\n\n\[ \n\\mathop{\\prod }\\limits_{{p \\leq {2m}}}p = \\mathop{\\prod }\\limits_{{p \\leq {2m} - 1}}p < {4}^{{2m} - 1}\n\]\n\nby the induction hy... | Yes |
Theorem 2. \( \pi \left( n\right) < \frac{cn}{\log n} \). | Clearly\n\n\[ \n{4}^{n} > \mathop{\prod }\limits_{{p \leq n}}p > \mathop{\prod }\limits_{{\sqrt{n} \leq p \leq n}}p > {\sqrt{n}}^{\pi \left( n\right) - \pi \left( \sqrt{n}\right) } \n\]\n\nTaking logarithms we obtain\n\n\[ \nn\log 4 > \left( {\pi \left( n\right) - \pi \left( \sqrt{n}\right) }\right) \frac{1}{2}\log n \... | Yes |
Theorem 3. \( \pi \left( n\right) > \frac{cn}{\log n} \) . | For this we use Lemmas 6 and 2. From these we obtain\n\n\[ \n{\left( 2n\right) }^{\pi \left( {2n}\right) } > \left( \begin{matrix} {2n} \\ n \end{matrix}\right) > \frac{{4}^{n}}{2n}\n\]\n\nTaking logarithms, we find that\n\n\[ \n\left( {\pi \left( n\right) + 1}\right) \log {2n} > \log \left( {2}^{2n}\right) = {2n}\log ... | Yes |
Theorem 4. \( R\left( x\right) = \mathop{\sum }\limits_{{p \leq x}}\frac{1}{p} = \log \log x + O\left( 1\right) \) . | In fact\n\n\[ R\left( x\right) = \mathop{\sum }\limits_{{n = 2}}^{x}\frac{S\left( n\right) - S\left( {n - 1}\right) }{\log n} \]\n\n\[ = \mathop{\sum }\limits_{{n = 2}}^{x}S\left( n\right) \left( {\frac{1}{\log n} - \frac{1}{\log \left( {n + 1}\right) }}\right) + O\left( 1\right) \]\n\n\[ = \mathop{\sum }\limits_{{n = ... | Yes |
Theorem 5. If \( \pi \left( x\right) \sim \frac{cx}{\log x} \), then \( c = 1 \) . | Since\n\n\[ R\left( x\right) = \mathop{\sum }\limits_{{n = 1}}^{x}\frac{\pi \left( n\right) - \pi \left( {n - 1}\right) }{n} \]\n\n\[ = \mathop{\sum }\limits_{{n = 1}}^{x}\frac{\pi \left( n\right) }{{n}^{2}} + O\left( 1\right) \]\n\n\( \pi \left( n\right) \sim \frac{cx}{\log x} \) would imply\n\n\[ \mathop{\sum }\limit... | No |
Theorem 6. For every integer \( r \) there exists a prime \( p \) with\n\n\[ 3 \cdot {2}^{{2r} - 1} < p < 3 \cdot {2}^{2r}. \]\n | We restate several of our lemmas in the form in which they will be used.\n\n(1) If \( n < p < {2n} \) then \( p \) occurs exactly once in \( \left( \begin{matrix} {2n} \\ n \end{matrix}\right) \) .\n\n(2) If \( 2 \cdot {2}^{{2r} - 1} < p < 3 \cdot {2}^{{2r} - 1} \) then \( p \) does not occur in \( \left( \begin{matrix... | Yes |
Theorem 2. \( \\left\\{ {{p}^{n}\\left( c\\right) }\\right\\}, n \\geq 0 \\), is relatively prime for all \( c \) if and only if \( p\\left( x\\right) \) belongs to one of the following six classes of polynomials. | Proof. In view of (3) we need only verify that the particular solutions yield the six sequences given above. | No |
Lemma 1. If \( p \) is a prime and \( {\varepsilon }_{n}\left( p\right) \) is defined by \( {\varepsilon }_{n}\left( p\right) = - 1 \) when \( \left( {p - 1}\right) \mid n \) and \( {\varepsilon }_{n}\left( p\right) = 0 \) when \( \left( {p - 1}\right) \) does not divide \( n \) then\n\n\[ \n{S}_{n}\left( p\right) \equ... | A simple proof of (2) is given in [2, p. 90]. | No |
In the chapter opener we mentioned briefly how Albert Einstein showed that a limit exists to how fast any object can travel. Given Einstein's equation for the mass of a moving object, what is the value of this bound? | Our starting point is Einstein's equation for the mass of a moving object,\n\n\[ m = \frac{{m}_{0}}{\sqrt{1 - \frac{{v}^{2}}{{c}^{2}}}} \]\n\nwhere \( {m}_{0} \) is the object’s mass at rest, \( v \) is its speed, and \( c \) is the speed of light. To see how the mass changes at high speeds, we can graph the ratio of m... | Yes |
a. What is the instantaneous velocity of the ball when it hits the ground? | The first thing to do is determine how long it takes the ball to reach the ground. To do this, set \( s\left( t\right) = 0 \) . Solving\n\n\( - {16}{t}^{2} + {64} = 0 \), we get \( t = 2 \), so it take 2 seconds for the ball to reach the ground.\n\nThe instantaneous velocity of the ball as it strikes the ground is \( v... | Yes |
a. Find the force on the face of the dam when the reservoir is full. | We begin by establishing a frame of reference. As usual, we choose to orient the \( x \) -axis vertically, with the downward direction being positive. This time, however, we are going to let \( x = 0 \) represent the top of the dam, rather than the surface of the water. When the reservoir is full, the surface of the wa... | Yes |
a. Evaluate \( f\left( 2\right) \). b. Solve \( f\left( x\right) = 4 \) . | a. To evaluate \( f\left( 2\right) \), locate the point on the curve where \( x = 2 \), then read the \( y \) -coordinate of that point. The point has coordinates \( \left( {2,1}\right) \), so \( f\left( 2\right) = 1 \) . See Figure 8.\n\nb. To solve \( f\left( x\right) = 4 \), we find the output value 4 on the vertica... | Yes |
Find the domain of the following function: \( \{ \left( {2,{10}}\right) ,\left( {3,{10}}\right) ,\left( {4,{20}}\right) ,\left( {5,{30}}\right) ,\left( {6,{40}}\right) \} \) . | Solution First identify the input values. The input value is the first coordinate in an ordered pair. There are no restrictions, as the ordered pairs are simply listed. The domain is the set of the first coordinates of the ordered pairs.\n\n\( \{ 2,3,4,5,6\} \) | Yes |
Find the domain of the function \( f\left( x\right) = \frac{x + 1}{2 - x} \) . | Solution When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for \( x \) .\n\n\[ 2 - x = 0 \]\n\n\[ - x = - 2 \]\n\n\[ x = 2 \]\n\nNow, we will exclude 2 from the domain. The answers are all real numb... | Yes |
Find the domain and range of the function \( f \) whose graph is shown in Figure 11. | Solution The input quantity along the horizontal axis is \ | No |
Find the domain and range of \( f\left( x\right) = 2\sqrt{x + 4} \) . | Solution We cannot take the square root of a negative number, so the value inside the radical must be nonnegative.\n\n\[ x + 4 \geq 0\text{when}x \geq - 4 \]\n\nThe domain of \( f\left( x\right) \) is \( \lbrack - 4,\infty ) \).\n\nWe then find the range. We know that \( f\left( {-4}\right) = 0 \), and the function val... | Yes |
Given the function \( g\left( t\right) \) shown in Figure 1, find the average rate of change on the interval \( \left\lbrack {-1,2}\right\rbrack \) . | Solution At \( t = - 1 \), Figure 2 shows \( g\left( {-1}\right) = 4 \) . At \( t = 2 \), the graph shows \( g\left( 2\right) = 1 \) .\n\n\n\nFigure 2\n\nThe horizontal change \( {\Delta t} = 3 \) is shown by the red a... | Yes |
Using Figure 1, evaluate \( f\left( {g\left( 1\right) }\right) \). | Solution To evaluate \( f\left( {g\left( 1\right) }\right) \), we start with the inside evaluation. See Figure 2.\n\n\n\nFigure 2\n\nWe evaluate \( g\left( 1\right) \) using the graph of \( g\left( x\right) \), finding... | Yes |
Given \( f\left( t\right) = {t}^{2} - t \) and \( h\left( x\right) = {3x} + 2 \), evaluate \( f\left( {h\left( 1\right) }\right) \). | Solution Because the inside expression is \( h\left( 1\right) \), we start by evaluating \( h\left( x\right) \) at 1 . \n\n\[ \nh\left( 1\right) = 3\left( 1\right) + 2 \n\] \n\n\[ \nh\left( 1\right) = 5 \n\] \n\nThen \( f\left( {h\left( 1\right) }\right) = f\left( 5\right) \), so we evaluate \( f\left( t\right) \) at a... | Yes |
Find the domain of\n\n\\[ \n\\left( {f \\circ g}\\right) \\left( x\\right) \\text{where}f\\left( x\\right) = \\frac{5}{x - 1}\\text{and}g\\left( x\\right) = \\frac{4}{{3x} - 2}\n\\]\n | Solution The domain of \\( g\\left( x\\right) \\) consists of all real numbers except \\( x = \\frac{2}{3} \\), since that input value would cause us to divide by 0 . Likewise, the domain of \\( f \\) consists of all real numbers except 1 . So we need to exclude from the domain of \\( g\\left( x\\right) \\) that value ... | Yes |
Given \( f\left( x\right) = \left| x\right| \), sketch a graph of \( h\left( x\right) = f\left( {x + 1}\right) - 3 \) . | Solution The function \( f \) is our toolkit absolute value function. We know that this graph has a V shape, with the point at the origin. The graph of \( h \) has transformed \( f \) in two ways: \( f\left( {x + 1}\right) \) is a change on the inside of the function, giving a horizontal shift left by 1, and the subtra... | Yes |
Example 10 Reflecting a Tabular Function Horizontally and Vertically\n\nA function \( f\left( x\right) \) is given as Table 6. Create a table for the functions below.\n\na. \( g\left( x\right) = - f\left( x\right) \; \) b. \( h\left( x\right) \) | ## Solution\n\na. For \( g\left( x\right) \), the negative sign outside the function indicates a vertical reflection, so the \( x \) -values stay the same and each output value will be the opposite of the original output value. See Table 7.\n\n<table><tr><td>\( x \)</td><td>2</td><td>4</td><td>6</td><td>8</td></tr><tr>... | "No" |
Is the function \( f\left( x\right) = {x}^{3} + {2x} \) even, odd, or neither? | Solution Without looking at a graph, we can determine whether the function is even or odd by finding formulas for the reflections and determining if they return us to the original function. Let’s begin with the rule for even functions.\n\n\[ f\left( {-x}\right) = {\left( -x\right) }^{3} + 2\left( {-x}\right) = - {x}^{3... | Yes |
Solve \( \left| {x - 5}\right| < 4 \) . | Solution With both approaches, we will need to know first where the corresponding equality is true. In this case we first will find where \( \left| {x - 5}\right| = 4 \) . We do this because the absolute value is a function with no breaks, so the only way the function values can switch from being less than \( 4 \) to b... | Yes |
Given the function \( f\left( x\right) = - \frac{1}{2}\left| {{4x} - 5}\right| + 3 \), determine the \( x \) -values for which the function values are negative. | Solution We are trying to determine where \( f\left( x\right) < 0 \), which is when \( - \frac{1}{2}\left| {{4x} - 5}\right| + 3 < 0 \) . We begin by isolating the absolute value.\n\n\[ - \frac{1}{2}\left| {{4x} - 5}\right| < - 3\;\text{Multiply both sides by -2, and reverse the inequality.}\]\n\n\[ \left| {{4x} - 5}\r... | Yes |
If \( f\left( x\right) = \frac{1}{x + 2} \) and \( g\left( x\right) = \frac{1}{x} - 2 \), is \( g = {f}^{-1} \) ? | \[ g\left( {f\left( x\right) }\right) = \frac{1}{\left( \frac{1}{x + 2}\right) } - 2 \]\n\[ = x + 2 - 2 \]\n\[ = x \]\nso\n\[ g = {f}^{-1}\text{ and }f = {g}^{-1} \]\nThis is enough to answer yes to the question, but we can also verify the other formula.\n\[ f\left( {g\left( x\right) }\right) = \frac{1}{\frac{1}{x} - 2... | Yes |
Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature. | \[ C = \frac{5}{9}\left( {F - {32}}\right) \] \n\n\[ C \cdot \frac{9}{5} = F - {32} \] \n\n\[ F = \frac{9}{5}C + {32} \] \n\nBy solving in general, we have uncovered the inverse function. If \n\n\[ C = h\left( F\right) = \frac{5}{9}\left( {F - {32}}\right) \] \n\nthen \n\n\[ F = {h}^{-1}\left( C\right) = \frac{9}{5}C +... | Yes |
Given the graph of \( f\left( x\right) \) in Figure 9, sketch a graph of \( {f}^{-1}\left( x\right) \) . | Solution This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of \( \left( {0,\infty }\right) \) and range of \( \left( {-\infty ,\infty }\right) \), so the inverse will have a domain of \( \left( {-\infty ,\infty }\right) \) and range of \( \left( {0,... | Yes |
Write the point-slope form of an equation of a line with a slope of 3 that passes through the point \( \left( {6, - 1}\right) \) . Then rewrite it in the slope-intercept form. | Solution Let’s figure out what we know from the given information. The slope is 3, so \( m = \) 3. We also know one point, so we know \( {x}_{1} = 6 \) and \( {y}_{1} = - 1 \) . Now we can substitute these values into the general point-slope equation.\n\n\[ y - {y}_{1} = m\left( {x - {x}_{1}}\right) \]\n\n\[ y - \left(... | Yes |
Graph \( f\left( x\right) = - \frac{2}{3}x + 5 \) by plotting points. | Solution Begin by choosing input values. This function includes a fraction with a denominator of 3, so let’s choose multiples of 3 as input values. We will choose 0,3 , and 6 .\n\nEvaluate the function at each input value, and use the output value to identify coordinate pairs.\n\n\[ x = 0\;f\left( 0\right) = - \frac{2}... | Yes |
A line passes through the points \( \left( {-2,6}\right) \) and \( \left( {4,5}\right) \) . Find the equation of a perpendicular line that passes through the point \( \left( {4,5}\right) \) . | Solution From the two points of the given line, we can calculate the slope of that line.\n\n\[ \n{m}_{1} = \frac{5 - 6}{4 - \left( {-2}\right) } \n\]\n\n\[ \n= \frac{-1}{6} \n\]\n\n\[ \n= - \frac{1}{6} \n\]\n\nFind the negative reciprocal of the slope.\n\n\[ \n{m}_{2} = \frac{-1}{-\frac{1}{6}} \n\]\n\n\[ \n= - 1\left( ... | Yes |
Find the least squares regression line using the cricket-chirp data in Table 1. | 1. Enter the input (chirps) in List 1 (L1).\n2. Enter the output (temperature) in List 2 (L2). See Table 2.\n<table><tr><td>L1</td><td>44</td><td>35</td><td>20.4</td><td>33</td><td>31</td><td>35</td><td>18.5</td><td>37</td><td>26</td></tr><tr><td>L2</td><td>80.5</td><td>70.5</td><td>57</td><td>66</td><td>68</td><td>72<... | Yes |
Express \( \sqrt{-9} \) in standard form. | Solution \( \sqrt{-9} = \sqrt{9}\sqrt{-1} = {3i} \)\n\nIn standard form, this is \( 0 + {3i} \) . | Yes |
Plot the complex number \( 3 - {4i} \) on the complex plane. | Solution The real part of the complex number is 3, and the imaginary part is \( - {4i} \). We plot the ordered pair \( \left( {3, - 4}\right) \) as shown in Figure 3. | Yes |
Add \( 3 - {4i} \) and \( 2 + {5i} \) . | Solution We add the real parts and add the imaginary parts.\n\n\[ \left( {a + {bi}}\right) + \left( {c + {di}}\right) = \left( {a + c}\right) + \left( {b + d}\right) i \]\n\n\[ \left( {3 - {4i}}\right) + \left( {2 + {5i}}\right) = \left( {3 + 2}\right) + \left( {-4 + 5}\right) i \]\n\n\[ = 5 + i \] | Yes |
Multiply \( \left( {4 + {3i}}\right) \left( {2 - {5i}}\right) \). | Solution Use \( \left( {a + {bi}}\right) \left( {c + {di}}\right) = \left( {{ac} - {bd}}\right) + \left( {{ad} + {bc}}\right) i \)\n\n\[ \left( {4 + {3i}}\right) \left( {2 - {5i}}\right) = \left( {4 \cdot 2 - 3 \cdot \left( {-5}\right) }\right) + \left( {4 \cdot \left( {-5}\right) + 3 \cdot 2}\right) i \]\n\n\[ = \left... | Yes |
Let \( f\left( x\right) = \frac{2 + x}{x + 3} \) . Evaluate \( f\left( {10i}\right) \) . | Solution Substitute \( x = {10i} \) and simplify.\n\n\[ \frac{2 + {10i}}{{10i} + 3} \]\nSubstitute \( {10i} \) for \( x \) .\n\n\[ \frac{2 + {10i}}{3 + {10i}} \]\nRewrite the denominator in standard form.\n\n\[ \frac{2 + {10i}}{3 + {10i}} \cdot \frac{3 - {10i}}{3 - {10i}} \]\nPrepare to multiply the numerator and denom... | Yes |
Evaluate \( {i}^{35} \) . | Solution Since \( {i}^{4} = 1 \), we can simplify the problem by factoring out as many factors of \( {i}^{4} \) as possible. To do so, first determine how many times 4 goes into 35 : \( {35} = 4 \cdot 8 + 3 \) .\n\n\[ \n{i}^{35} = {i}^{4 \cdot 8 + 3} = {i}^{4 \cdot 8} \cdot {i}^{3} = {\left( {i}^{4}\right) }^{8} \cdot ... | Yes |
Find the domain and range of \( f\left( x\right) = - 5{x}^{2} + {9x} - 1 \) . | Solution As with any quadratic function, the domain is all real numbers.\n\nBecause \( a \) is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the \( x \) -value of the vertex.\n\n\[ h = - \frac{b}{2a} \]\n\n\[ = - \frac{9}{2\left( {-5}\righ... | Yes |
Find the \( y \) - and \( x \) -intercepts of the quadratic \( f\left( x\right) = 3{x}^{2} + {5x} - 2 \) . | Solution We find the \( y \) -intercept by evaluating \( f\left( 0\right) \).\n\n\[ f\left( 0\right) = 3{\left( 0\right) }^{2} + 5\left( 0\right) - 2 \]\n\n\[ = - 2 \]\n\nSo the \( y \) -intercept is at \( \left( {0, - 2}\right) \).\n\nFor the \( x \) -intercepts, we find all solutions of \( f\left( x\right) = 0 \).\n\... | Yes |
Solve \( {x}^{2} + x + 2 = 0 \) . | Solution Let’s begin by writing the quadratic formula: \( x = \frac{-b \pm \sqrt{{b}^{2} - {4ac}}}{2a} \) .\n\nWhen applying the quadratic formula, we identify the coefficients \( a, b \) and \( c \) . For the equation \( {x}^{2} + x + 2 = 0 \), we have \( a = 1, b = 1 \), and \( c = 2 \) . Substituting these values in... | Yes |
Describe the end behavior of the graph of \( f\left( x\right) = - {x}^{9} \) . | The exponent of the power function is 9 (an odd number). Because the coefficient is -1 (negative), the graph is the reflection about the \( x \) -axis of the graph of \( f\left( x\right) = {x}^{9} \) . Figure 6 shows that as \( x \) approaches infinity, the output decreases without bound. As \( x \) approaches negative... | Yes |
Given the polynomial function \( f\left( x\right) = \left( {x - 2}\right) \left( {x + 1}\right) \left( {x - 4}\right) \), written in factored form for your convenience, determine the \( y \) - and \( x \) -intercepts. | Solution The \( y \) -intercept occurs when the input is zero so substitute 0 for \( x \) .\n\n\[ f\left( 0\right) = \left( {0 - 2}\right) \left( {0 + 1}\right) \left( {0 - 4}\right) \]\n\n\[ = \left( {-2}\right) \left( 1\right) \left( {-4}\right) \]\n\n\[ = 8 \]\n\nThe \( y \) -intercept is \( \left( {0,8}\right) \).\... | Yes |
Given the polynomial function \( f\left( x\right) = {x}^{4} - 4{x}^{2} - {45} \), determine the \( y \) - and \( x \) -intercepts. | Solution The \( y \) -intercept occurs when the input is zero.\n\n\[ f\left( 0\right) = {\left( 0\right) }^{4} - 4{\left( 0\right) }^{2} - {45} \]\n\n\[ = - {45} \]\n\nThe \( y \) -intercept is \( \left( {0, - {45}}\right) \).\n\nThe \( x \) -intercepts occur when the output is zero. To determine when the output is zer... | Yes |
Find the \( x \) -intercepts of \( f\left( x\right) = {x}^{3} - 5{x}^{2} - x + 5 \) . | Solution Find solutions for \( f\left( x\right) = 0 \) by factoring.\n\n\[ \n{x}^{3} - 5{x}^{2} - x + 5 = 0\;\text{Factor by grouping.} \]\n\n\( {x}^{2}\left( {x - 5}\right) - \left( {x - 5}\right) = 0\; \) Factor out the common factor.\n\n\( \left( {{x}^{2} - 1}\right) \left( {x - 5}\right) = 0\; \) Factor the differe... | Yes |
Find the \( y \) - and \( x \) -intercepts of \( g\left( x\right) = {\left( x - 2\right) }^{2}\left( {{2x} + 3}\right) \) . | Solution The \( y \) -intercept can be found by evaluating \( g\left( 0\right) \) .\n\n\[ g\left( 0\right) = {\left( 0 - 2\right) }^{2}\left( {2\left( 0\right) + 3}\right) \]\n\n\[ = {12} \]\n\nSo the \( y \) -intercept is \( \left( {0,{12}}\right) \) .\n\nThe \( x \) -intercepts can be found by solving \( g\left( x\ri... | Yes |
Find the \( x \) -intercepts of \( h\left( x\right) = {x}^{3} + 4{x}^{2} + x - 6 \) . | Solution This polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing... | Yes |
Find the maximum number of turning points of each polynomial function.\na. \( f\left( x\right) = - {x}^{3} + 4{x}^{5} - 3{x}^{2} + 1\; \) b. \( f\left( x\right) = - {\left( x - 1\right) }^{2}\left( {1 + 2{x}^{2}}\right) \) | a. \( f\left( x\right) = - {x}^{3} + 4{x}^{5} - 3{x}^{2} + 1 \)\nFirst, rewrite the polynomial function in descending order: \( f\left( x\right) = 4{x}^{5} - {x}^{3} - 3{x}^{2} + 1 \)\nIdentify the degree of the polynomial function. This polynomial function is of degree 5 .\nThe maximum number of turning points is \( 5... | Yes |
Sketch a graph of \( f\left( x\right) = - 2{\left( x + 3\right) }^{2}\left( {x - 5}\right) \) . | Solution This graph has two \( x \) -intercepts. At \( x = - 3 \), the factor is squared, indicating a multiplicity of 2 . The graph will bounce at this \( x \) -intercept. At \( x = 5 \), the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept.\n\nThe \( y \) -interce... | Yes |
Divide \( 5{x}^{2} + {3x} - 2 \) by \( x + 1 \) . | \[ x + 1\overset{―}{)5{x}^{2} + {3x} - 2}\;\text{Set up division problem.} \]\n\[ x + 1)\frac{5x}{5{x}^{2} + {3x} - 2}\;5{x}^{2}\text{divided by}x\text{is}{5x}. \]\n\[ x + 1\overset{5x}{\overline{5{x}^{2} + {3x} - 2}}\;\text{ Multiply }x + 1\text{ by }{5x}. \]\n\[ \begin{array}{l} x + 1)\frac{{5x} - 2}{5{x}^{2} + {3x} ... | Yes |
Use synthetic division to divide \( 5{x}^{2} - {3x} - {36} \) by \( x - 3 \) . | Solution Begin by setting up the synthetic division. Write \( k \) and the coefficients.\n\n\[ \begin{array}{llll} 3 & 5 & - 3 & - {36} \end{array} \]\n\nBring down the lead coefficient. Multiply the lead coefficient by \( k \).  \) is a factor of \( {x}^{3} - 6{x}^{2} - x + {30} \) . Find the remaining factors. Use the factors to determine the zeros of the polynomial. | Solution We can use synthetic division to show that \( \left( {x + 2}\right) \) is a factor of the polynomial.\n\n\n\nThe remainder is zero, so \( \left( {x + 2}\right) \) is a factor of the polynomial. We can use th... | Yes |
List all possible rational zeros of \( f\left( x\right) = 2{x}^{4} - 5{x}^{3} + {x}^{2} - 4 \) . | The only possible rational zeros of \( f\left( x\right) \) are the quotients of the factors of the last term,-4, and the factors of the leading coefficient, 2.\n\nThe constant term is -4 ; the factors of -4 are \( p = \pm 1, \pm 2, \pm 4 \) .\n\nThe leading coefficient is 2 ; the factors of 2 are \( q = \pm 1, \pm 2 \)... | Yes |
Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any. | Shifting the graph left 2 and up 3 would result in the function\n\n\[ f\left( x\right) = \frac{1}{x + 2} + 3 \]\n\nor equivalently, by giving the terms a common denominator,\n\n\[ f\left( x\right) = \frac{{3x} + 7}{x + 2} \]\n\nThe graph of the shifted function is displayed in Figure 7.\n\n![f7473d67-b7fd-4fb3-af38-0b0... | Yes |
Show that \( f\left( x\right) = \frac{1}{x + 1} \) and \( {f}^{-1}\left( x\right) = \frac{1}{x} - 1 \) are inverses, for \( x \neq 0, - 1 \) . | Solution We must show that \( {f}^{-1}\left( {f\left( x\right) }\right) = x \) and \( f\left( {{f}^{-1}\left( x\right) }\right) = x \) .\n\n\[ \n{f}^{-1}\left( {f\left( x\right) }\right) = {f}^{-1}\left( \frac{1}{x + 1}\right) \n\]\n\n\[ \n= \frac{1}{\frac{1}{x + 1}} - 1 \n\]\n\n\[ \n= \left( {x + 1}\right) - 1 \n\]\n\... | Yes |
Restrict the domain and then find the inverse of\n\n\[ f\left( x\right) = {\left( x - 2\right) }^{2} - 3. \] | Solution We can see this is a parabola with vertex at \( \left( {2, - 3}\right) \) that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to \( x \geq 2 \) .\n\nTo... | Yes |
Find the domain of the function \( f\left( x\right) = \sqrt{\frac{\left( {x + 2}\right) \left( {x - 3}\right) }{\left( x - 1\right) }} \) . | Because a square root is only defined when the quantity under the radical is non-negative, we need to determine where \( \frac{\left( {x + 2}\right) \left( {x - 3}\right) }{\left( x - 1\right) } \geq 0 \) . The output of a rational function can change signs (change from positive to negative or vice versa) at \( x \) -i... | Yes |
Example 8 Finding the Inverse of a Rational Function\n\nThe function \( C = \frac{{20} + {0.4n}}{{100} + n} \) represents the concentration \( C \) of an acid solution after \( n\mathrm{\;{mL}} \) of \( {40}\% \) solution has been\n\nadded to \( {100}\mathrm{\;{mL}} \) of a \( {20}\% \) solution. First, find the invers... | Solution We first want the inverse of the function. We will solve for \( n \) in terms of \( C \) .\n\n\[ C = \frac{{20} + {0.4n}}{{100} + n} \]\n\n\[ C\left( {{100} + n}\right) = {20} + {0.4n} \]\n\n\[ {100}\mathrm{C} + \mathrm{C}n = {20} + {0.4n} \]\n\n\[ {100C} - {20} = {0.4n} - {Cn} \]\n\n\[ {100C} - {20} = \left( ... | Yes |
A quantity \( y \) varies inversely with the cube of \( x \) . If \( y = {25} \) when \( x = 2 \), find \( y \) when \( x \) is 6 . | Solution The general formula for inverse variation with a cube is \( y = \frac{k}{{x}^{3}} \) . The constant can be found by multiplying\n\n\( y \) by the cube of \( x \) .\n\[ k = {x}^{3}y \]\n\[ = {2}^{3} \cdot {25} \]\n\[ = {200} \]\n\nNow we use the constant to write an equation that represents this relationship.\n... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.