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Proposition 11.6.7. Suppose \( \left( {X, d}\right) \) is a compact metric space, \( {f}_{n} \in C\left( {X,\mathbb{C}}\right) \), and \( \left\{ {f}_{n}\right\} \) converges uniformly, then \( \left\{ {f}_{n}\right\} \) is uniformly equicontinuous. | Proof. Let \( \varepsilon > 0 \) be given. As \( \left\{ {f}_{n}\right\} \) converges uniformly, there is an \( N \in \mathbb{N} \) such that for all \( n \geq N \)\n\n\[ \left| {{f}_{n}\left( x\right) - {f}_{N}\left( x\right) }\right| < \varepsilon /3\;\text{ for all }x \in X. \]\n\nAs \( X \) is compact, any continuo... | Yes |
Proposition 11.6.8. A compact metric space \( \left( {X, d}\right) \) contains a countable dense subset, that is, there exists a countable \( D \subset X \) such that \( \bar{D} = X \) . | Proof. For each \( n \in \mathbb{N} \) there are finitely many balls of radius \( 1/n \) that cover \( X \) (as \( X \) is compact). That is, for every \( n \), there exists a finite set of points \( {x}_{n,1},{x}_{n,2},\ldots ,{x}_{n,{k}_{n}} \) such that\n\n\[ X = \mathop{\bigcup }\limits_{{j = 1}}^{{k}_{n}}B\left( {... | Yes |
Corollary 11.6.10. Let \( \left( {X, d}\right) \) be a compact metric space. Let \( S \subset C\left( {X,\mathbb{C}}\right) \) be a closed, bounded and uniformly equicontinuous set. Then \( S \) is compact. | The theorem says that \( S \) is sequentially compact and that means compact in a metric space. Recall that the closed unit ball in \( C\left( {\left\lbrack {0,1}\right\rbrack ,\mathbb{R}}\right) \) (and therefore also in \( C\left( {\left\lbrack {0,1}\right\rbrack ,\mathbb{C}}\right) \) ) is not compact. Hence it cann... | No |
Corollary 11.6.11. Suppose \( \left\{ {f}_{n}\right\} \) is a sequence of differentiable functions on \( \left\lbrack {a, b}\right\rbrack ,\left\{ {f}_{n}^{\prime }\right\} \) is uniformly bounded, and there is an \( {x}_{0} \in \left\lbrack {a, b}\right\rbrack \) such that \( \left\{ {{f}_{n}\left( {x}_{0}\right) }\ri... | Proof. The trick is to use the mean value theorem. If \( M \) is the uniform bound on \( \left\{ {f}_{n}^{\prime }\right\} \), then by the mean value theorem for any \( n \)\n\n\[ \left| {{f}_{n}\left( x\right) - {f}_{n}\left( y\right) }\right| \leq M\left| {x - y}\right| \;\text{ for all }x, y \in X. \]\n\nAll the \( ... | Yes |
Theorem 11.7.1 (Weierstrass approximation theorem). If \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{C} \) is continuous, then there exists a sequence \( \left\{ {p}_{n}\right\} \) of polynomials converging to \( f \) uniformly on \( \left\lbrack {a, b}\right\rbrack \) . Furthermore, if \( f \) is real-va... | Proof. For \( x \in \left\lbrack {0,1}\right\rbrack \) define\n\n\[ g\left( x\right) \mathrel{\text{:=}} f\left( {\left( {b - a}\right) x + a}\right) - f\left( a\right) - x\left( {f\left( b\right) - f\left( a\right) }\right) .\n\]\n\nIf we prove the theorem for \( g \) and find the sequence \( \left\{ {p}_{n}\right\} \... | Yes |
Corollary 11.7.2. The metric space \( C\left( {\left\lbrack {a, b}\right\rbrack ,\mathbb{C}}\right) \) contains a countable dense subset. | Proof. Without loss of generality suppose that we are dealing with \( C\left( {\left\lbrack {a, b}\right\rbrack ,\mathbb{R}}\right) \) (why?). The real polynomials are dense in \( C\left( {\left\lbrack {a, b}\right\rbrack ,\mathbb{R}}\right) \) by Weierstrass. If we show that any real polynomial can be approximated by ... | Yes |
Corollary 11.7.4. Let \( \left\lbrack {-a, a}\right\rbrack \) be an interval. Then there is a sequence of real polynomials \( \left\{ {p}_{n}\right\} \) that converges uniformly to \( \left| x\right| \) on \( \left\lbrack {-a, a}\right\rbrack \) and such that \( {p}_{n}\left( 0\right) = 0 \) for all \( n \) . | Proof. As \( f\left( x\right) \mathrel{\text{:=}} \left| x\right| \) is continuous and real-valued on \( \left\lbrack {-a, a}\right\rbrack \), the Weierstrass theorem gives a sequence of real polynomials \( \left\{ {\widetilde{p}}_{n}\right\} \) that converges to \( f \) uniformly on \( \left\lbrack {-a, a}\right\rbrac... | Yes |
Proposition 11.7.11. Suppose \( \mathcal{A} \) is an algebra of complex-valued functions on a set \( X \), that separates points and vanishes at no point. Suppose \( x, y \) are distinct points of \( X \), and \( c, d \in \mathbb{C} \) . Then there is an \( f \in \mathcal{A} \) such that\n\n\[ f\left( x\right) = c,\;f\... | Proof. There must exist an \( g, h, k \in \mathcal{A} \) such that\n\n\[ g\left( x\right) \neq g\left( y\right) ,\;h\left( x\right) \neq 0,\;k\left( y\right) \neq 0. \]\n\nLet\n\n\[ f \mathrel{\text{:=}} c\frac{\left( {g - g\left( y\right) }\right) h}{\left( {g\left( x\right) - g\left( y\right) }\right) h\left( x\right... | Yes |
Theorem 11.7.16 (Stone-Weierstrass, complex version). Let \( X \) be a compact metric space and \( \mathcal{A} \) an algebra of complex-valued continuous functions on \( X \), such that \( \mathcal{A} \) separates points, vanishes at no point, and is self-adjoint. Then the closure \( \overline{\mathcal{A}} = C\left( {X... | Proof. Suppose \( {\mathcal{A}}_{\mathbb{R}} \subset \mathcal{A} \) is the set of the real-valued elements of \( \mathcal{A} \) . For any \( f \in \mathcal{A} \), write \( f = u + {iv} \) where \( u \) and \( v \) are real-valued. Then\n\n\[ u = \frac{f + \bar{f}}{2},\;v = \frac{f - \bar{f}}{2i}. \]\n\nSo \( u, v \in \... | Yes |
Proposition 11.8.1. A trigonometric polynomial \( f\left( x\right) = \mathop{\sum }\limits_{{n = - N}}^{N}{c}_{n}{e}^{inx} \) is real-valued for real \( x \) if and only if \( {c}_{-m} = \overline{{c}_{m}} \) for all \( m = - N,\ldots, N \) . | Proof. If \( f\left( x\right) \) is real-valued, that is \( \overline{f\left( x\right) } = f\left( x\right) \), then\n\n\[ \overline{{c}_{m}} = \overline{\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left( x\right) {e}^{-{imx}}{dx}} = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }\overline{f\left( x\right) {e}^{-{imx}}}{dx} = \frac{1}... | Yes |
Proposition 11.8.2. If\n\n\[ \mathop{\sum }\limits_{{n = - N}}^{N}{c}_{n}{e}^{inx} = 0 \] \n\nfor all \( x \in \left\lbrack {-\pi ,\pi }\right\rbrack \), then \( {c}_{n} = 0 \) for all \( n \) . | Proof. Proof follows immediately from the integral formula for \( {c}_{n} \) . | No |
Proposition 11.8.3. Let \( \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{c}_{n}{e}^{inx} \) be a Fourier series, and \( C,\alpha > 1 \) constants such that\n\n\[ \left| {c}_{n}\right| \leq \frac{C}{{\left| n\right| }^{\alpha }}\;\text{ for all }n \in \mathbb{Z} \smallsetminus \{ 0\} \]\n\nThen the series converges ... | The proof is to apply the Weierstrass \( M \) -test (Theorem 11.2.4) and the \( p \) -series test, to find that the series converges uniformly and hence to a continuous function. We can also take derivatives. | Yes |
Proposition 11.8.4. Let \( \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{c}_{n}{e}^{inx} \) be a Fourier series, and \( C,\alpha > 2 \) constants such that\n\n\[ \left| {c}_{n}\right| \leq \frac{C}{{\left| n\right| }^{\alpha }}\;\text{ for all }n \in \mathbb{Z} \smallsetminus \{ 0\} \]\n\nThen the series converges ... | The trick is to first notice that the series converges first to a continuous function by the previous proposition, so in particular it converges at some point. Then differentiate the partial sums\n\n\[ \mathop{\sum }\limits_{{n = - N}}^{N}{in}{c}_{n}{e}^{inx} \]\n\nand notice that\n\n\[ \left| {{in}{c}_{n}}\right| \leq... | Yes |
Theorem 11.8.7. Suppose \( f \) is a Riemann integrable function on \( \left\lbrack {a, b}\right\rbrack \) . Let \( \left\{ {\varphi }_{n}\right\} \) be an orthonormal system on \( \left\lbrack {a, b}\right\rbrack \) and suppose\n\n\[ f\left( x\right) \sim \mathop{\sum }\limits_{{n = 1}}^{\infty }{c}_{n}{\varphi }_{n}\... | Proof. Let us write\n\n\[ {\int }_{a}^{b}{\left| f - {p}_{n}\right| }^{2} = {\int }_{a}^{b}{\left| f\right| }^{2} - {\int }_{a}^{b}f\overline{{p}_{n}} - {\int }_{a}^{b}\bar{f}{p}_{n} + {\int }_{a}^{b}{\left| {p}_{n}\right| }^{2}. \]\n\nNow\n\n\[ {\int }_{a}^{b}f\overline{{p}_{n}} = {\int }_{a}^{b}f\mathop{\sum }\limits... | Yes |
Corollary 11.8.10. Let \( f \) be a \( {2\pi } \) -periodic function Riemann integrable on \( \left\lbrack {-\pi ,\pi }\right\rbrack \) . Suppose there exist \( x \in \mathbb{R} \) and \( \delta > 0 \) such that \( f \) is continuous piecewise smooth on \( \left\lbrack {x - \delta, x + \delta }\right\rbrack \), then\n\... | The proof of the corollary is left as an exercise. | No |
Corollary 11.8.11. Suppose \( f \) is a \( {2\pi } \) -periodic function, Riemann integrable on \( \left\lbrack {-\pi ,\pi }\right\rbrack \) . If \( J \) is an open interval and \( f\left( x\right) = 0 \) for all \( x \in J \), then \( \lim {s}_{N}\left( {f;x}\right) = 0 \) for all \( x \in J \) . | To prove the first claim, take \( M = 0 \) in the theorem. The \ | No |
Theorem 11.8.12 (Parseval*). Let \( f \) and \( g \) be \( {2\pi } \) -periodic functions, Riemann integrable on \( \left\lbrack {-\pi ,\pi }\right\rbrack \) with\n\n\[ f\left( x\right) \sim \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{c}_{n}{e}^{inx}\;\text{ and }\;g\left( x\right) \sim \mathop{\sum }\limits_{{n ... | Proof. There exists (exercise) a continuous \( {2\pi } \) -periodic function \( h \) such that\n\n\[ \parallel f - h{\parallel }_{2} < \varepsilon \]\n\nVia Stone-Weierstrass, approximate \( h \) with a trigonometric polynomial uniformly. That is, there is a trigonometric polynomial \( P\left( x\right) \) such that \( ... | No |
\[ 2 \times \left( {3 + \left( {4 \times 2}\right) }\right) \text{.} \] | Solution Here we see a set of parentheses \ | No |
Evaluate\n\n\[ \left( {2 + \left( {3 \times \left( {4 + \left( {2 - 1}\right) }\right) - 1}\right) }\right) + 2 \] | Solution\n\n\[ \left( {2 + \left( {3 \times \left( {4 + \left( {2 - 1}\right) }\right) - 1}\right) }\right) + 2 = \left\lbrack {2 + \left( {3 \times \left\lbrack {4 + \left( {2 - 1}\right) }\right\rbrack - 1}\right) }\right\rbrack + 2 \]\n\n\[ = \left\lbrack {2 + \left( {3 \times \left\lbrack {4 + \left( 1\right) }\rig... | Yes |
Evaluate\n\n\[ \sqrt{{12} \times 3} - {20} \] | Solution\n\n\[ \sqrt{{12} \times 3} - {20} = \sqrt{\left( {12} \times 3\right) } - {20} \]\n\n\[ = \sqrt{\left( {36}\right) } - {20} \]\n\n\[ = 6 - {20} \]\n\n\[ = - {14} \]\n\nThis gives our final answer:\n\n\[ \sqrt{{12} \times 3} - {20} = - {14} \] | Yes |
Evaluate\n\n\[ 2 + {3}^{3} \] | Solution\n\n\[ 2 + {3}^{3} = 2 + {27} \]\n\n(1.9)\n\n\[ = {29} \]\n\nNotice we only cubed the 3 and not the expression \( 2 + 3 \), giving us a final answer of\n\n\[ 2 + {3}^{3} = {29} \] | Yes |
Evaluate\n\n\[ - {4}^{2} \] | \[ - {4}^{2} = - \left( {4}^{2}\right) \]\n\n\[ = - \left( {16}\right) \]\n\n\[ = - {16} \]\n\nHere, our final answer is\n\n\[ - {4}^{2} = - {16} \] | Yes |
Evaluate\n\n\[ 6 \div 2 \times 3 + 1 \times 8 \div 4 \] | \[ 6 \div 2 \times 3 + 1 \times 8 \div 4 = 3 \times 3 + 1 \times 8 \div 4 \]\n\n\[ = 9 + 1 \times 8 \div 4 \]\n\n\[ = 9 + 8 \div 4 \]\n\n\[ = 9 + 2 \]\n\n\[ = {11} \]\n\nSince this expression doesn't have any parentheses or exponents, we look for multiplication or division, starting on the left. First, we find \( 6 \di... | Yes |
Evaluate\n\n\[ 1 - 3 + 6 \] | Notes: Solution\n\n\[ 1 - 3 + 6 = - 2 + 6 \]\n\n(1.12)\n\n\[ = 4 \]\n\nBy doing addition and subtraction on the same level, from left to right, we get a final answer of\n\n\[ 1 - 3 + 6 = 4 \] | Yes |
Evaluate\n\n\[ - {2}^{2} + \sqrt{6 - 2} - 2\\left( {8 \\div 2 \\times \\left( {1 + 1}\\right) }\\right) \] | Solution Since this is more complicated than our earlier examples, let's make a table showing each step on the left, with an explanation on the right:\n\n<table><tr><td>\\( - {2}^{2} + \\sqrt{6 - 2} - 2\\left( {8 \\div 2 \\times \\left( {1 + 1}\\right) }\\right) \\) =</td><td>We have a bit of everything here, so let's ... | Yes |
Evaluate\n\n\[ \frac{1}{4} \times \frac{2}{3} \times 2 \] | lution With multiplication of fractions, we will work just like we do with any other type of real number and multiply from left to right. When multiplying two fractions together, we will multiply their numerators together (the tops) and we will multiply the denominators together (the bottoms).\n\n\[ \frac{1}{4} \times ... | Yes |
Evaluate\n\n\[ \n{\\left( \\frac{1 + 2}{5}\\right) }^{2} \n\] | Solution With exponentiation, we need to apply the exponent to both\nthe numerator and the denominator. This gives\n\n\[ \n{\\left( \\frac{1 + 2}{5}\\right) }^{2} = {\\left( \\frac{\\left( 1 + 2\\right) }{\\left( 5\\right) }\\right) }^{2} \n\]\n\n\[ \n= {\\left( \\frac{\\left( 3\\right) }{\\left( 5\\right) }\\right) }^... | Yes |
Evaluate\n\n\[ \frac{1}{2} - \frac{1}{3} + \frac{1}{4} \] | Since we only have addition and subtraction, we will work from left to right. This means that our first step is to subtract \( \frac{1}{3} \) from \( \frac{1}{2} \) . The denominators are different, so we don't yet have pieces that are all the same size. To make sure our pieces are all the same size, we will multiply e... | No |
Determine the value of \( f\left( {-2}\right) \) if \( f\left( x\right) = {x}^{2} + {4x} - {10} \) . | Solution First, we notice that the left side tells us that our input is \( x \) . Since we want to determine the value of \( f\left( {-2}\right) \), we’ll replace every \( x \) on the right side with \( \left( {-2}\right) \) .\n\n\[ f\left( {-2}\right) = {\left( -2\right) }^{2} + 4\left( {-2}\right) - {10} \]\n\n\[ = \... | Yes |
Find the difference quotient for \( g\left( t\right) = 2{t}^{2} - {3t} + 1 \) | Solution Here we have a function called \( g \), with \( t \) as its input. That means that in our difference quotient, we will have \( g \) instead of \( f \) and \( t \) instead\nof \( x \), but \( h \) will still be \( h \) . So, our difference quotient will look like\n\n\[ \frac{g\left( {t + h}\right) - g\left( t\r... | Yes |
Using \( f\left( x\right) \) and \( g\left( x\right) \) from above, determine \( j\left( x\right) = f\left( {g\left( x\right) }\right) \) . | Since \( g\left( x\right) \) is our input, we need to replace every \( x \) in \( f \) with ( \( x - 4 \) ). This gives us\n\n\[ j\left( x\right) = f\left( {g\left( x\right) }\right) = f\left( {x - 4}\right) \]\n\n\[ = 3{\left( x - 4\right) }^{2} \]\n\n\[ = 3\left( {x - 4}\right) \left( {x - 4}\right) \]\n\n(1.23)\n\n\... | Yes |
Using \( f\left( x\right) \) and \( g\left( x\right) \) from above, determine \( m\left( x\right) = f\left( {g\left( {g\left( x\right) }\right) }\right) \) . | Solution With multiple layers of composition, it's typically easiest to start on the inner layer first and then work your way out. Here the outermost function is \( f\left( x\right) \), then \( g\left( x\right) \) in the middle, and \( g\left( x\right) \) on the inside. We already know what \( g\left( x\right) \) looks... | Yes |
Evaluate \( \left( {{40} + 2}\right) \left( {{30} + 1}\right) \) . | Solution Typically, we would start this problem by looking at our order of operations. Our order of operations tells us to do everything inside the parentheses first, which would give us (42)(31), and then we would multiply these. However, we are going to use the distributive property instead. The distributive property... | Yes |
Factor the quadratic function \( f\left( x\right) = {x}^{2} + {5x} + 4 \) . | Solution As we noted above, the best starting point is to look for pairs of integers that we can multiply to get the constant term. We know that to get 4 , we could multiply any of the following pairs to get 4 :\n\n(A) 4 and 1 (C) -4 and -1\n\n(B) 2 and 2 (D) -2 and -2\n\nNow, we’ll look at the \( x \) term in the quad... | Yes |
Factor the quadratic function \( g\left( x\right) = {x}^{2} - {5x} + 6 \) . | Solution We'll start like we did in our last example by looking for pairs of integers that multiply to give us 6 :\n\n(A) 6 and 1 (C) -6 and -1\n\n(B) 3 and 2 (D) -3 and -2\n\nOut of these pairs, only -3 and -2 add to give us -5, the \( x \) coefficient in the quadratic. This tells us that the factors are \( x - 3 \) a... | Yes |
Factor the quadratic function \( h\left( x\right) = {x}^{2} - {7x} - {18} \) . | Solution We'll start like we did in our last example by looking for pairs of integers that multiply to give us -18:\n\n(A) -18 and 1 (D) -3 and 6\n\n(B) -9 and 2 (E) -2 and 9\n\n(C) -6 and 3 (F) -1 and 18\n\nOut of these pairs, only -9 and 2 add to give us -7, the \( x \) coefficient in the quadratic. This tells us tha... | Yes |
Factor the quadratic function \( m\left( x\right) = {x}^{2} + {3x} - {18} \) . | solution We'll start like we did in our last example by looking for pairs of integers that multiply to give us -18:\n\n(A) -18 and 1 (D) -3 and 6\n\n(B) -9 and 2 (E) -2 and 9\n\n(C) -6 and 3 (F) -1 and 18\n\nOut of these pairs, only -3 and 6 add to give us 3, the \( x \) coefficient in the quadratic. This tells us that... | No |
Factor the quadratic function \( h\left( t\right) = 6{t}^{2} - {7t} + 2 \) . | Here, our function has \( t \) instead of \( x \), but it really is in the form we need to use the quadratic formula; we'll just make sure to give the answer with \( t \) instead of \( x \) . We’ll find our roots, and then use those to help us find our factors. We'll start by identifying the values for a, b, and c, and... | Yes |
Find a factor of the function \( f\left( x\right) = {x}^{3} + 8{x}^{2} + {21x} + {18} \) . | Here, the constant term of the cubic is 18 , so we'll start by listing all of its factors, positive and negative. The factors are: 18, 9, 6, 3, 2, 1, -1, -2, -3, -6, -9, and -18 . There are a bunch, so as mentioned above, we'll start by checking the \ | No |
Completely factor \( f\left( t\right) = {t}^{3} + {t}^{2} - {4t} - 4 \) . | Solution We will start by grouping and then factoring each group:\n\n\[ f\left( t\right) = {t}^{3} + {t}^{2} - {4t} - 4 = \left( {{t}^{3} + {t}^{2}}\right) + \left( {-{4t} - 4}\right) \]\n\n\[ = {t}^{2}\left( {t + 1}\right) + \left( {-4}\right) \left( {t + 1}\right) \]\n\n\[ = {t}^{2}\left( {t + 1}\right) - 4\left( {t ... | Yes |
Factor \( g\left( t\right) = 8{t}^{3} - \frac{1}{27} \) completely. | Solution Here we see that \( g\left( t\right) \) has only two terms, and that one of them has \( {t}^{3} \) . This points me towards the last two patterns: they both have parts raised to the third power and only have two terms each. With \( g\left( t\right) \), we have subtraction, not addition, so this points us to th... | Yes |
Simplify \( {\left( \frac{{x}^{2}{y}^{4}}{x\sqrt{y}}\right) }^{2} \) | Anytime we simplify, we need to remember our order of operations. The order of operations tells us to start with terms that are inside of parentheses, so we will work on simplifying the fraction before we worry about the exponent on the outside. First, we will write everything using exponents rather than radicals so we... | Yes |
Simplify \( {\left( \sqrt{y} + \sqrt{x}\right) }^{2} \). | Solution We'll start again by focusing on the terms inside the parentheses and rewriting all radicals as exponents. This gives us\n\n\[ \n{\left( \sqrt{y} + \sqrt{x}\right) }^{2} = {\left( {y}^{1/2} + {x}^{1/2}\right) }^{2} \n\]\n\nThere is nothing that we can simplify inside the parentheses, so we now need to apply th... | Yes |
Evaluate \( {8}^{2/3} \). | Solution As first glance, this looks like we won't be able to do much with it. However, we can use our exponent rules to help us evaluate it. We can rewrite this as \( {\left( {8}^{2}\right) }^{1/3} \) or as \( {\left( {8}^{1/3}\right) }^{2} \) . We prefer the second version. With the first version we would have \( {\l... | Yes |
Write \( 5{\log }_{2}\left( x\right) + 3{\log }_{2}\left( {2y}\right) \) as a single logarithm. | Currently, this term is the sum of two logarithms, both with the same base, and we want to write it as a single logarithm. It looks like we may want to start with the second rule, \( {\log }_{b}\left( {xy}\right) = {\log }_{b}\left( x\right) + {\log }_{b}\left( y\right) \) . It looks like we are already in the form on ... | Yes |
Expand \( \ln \left( \frac{2{x}^{3}{y}^{3}}{w{z}^{5}}\right) \) into the sum and/or difference of multiple logarithms. | Solution Here, we want to rewrite as many simpler logarithms. First, we see that the logarithms has a quotient inside, so we can use the third rule to split it:\n\n\[ \ln \left( \frac{2{x}^{3}{y}^{3}}{w{z}^{5}}\right) = \ln \left( {2{x}^{3}{y}^{3}}\right) - \ln \left( {w{z}^{5}}\right) \]\n\nNow, we have products insid... | Yes |
Solve \( {5}^{{3x} - 1} - 2 = 0 \) for \( x \) . | Solution First, we will need to isolate the exponential term, \( {5}^{{3x} - 1} \) . Then, we will take log base 5 of both sides since the exponent has 5 as its base.\n\n\[ \n{5}^{{3x} - 1} - 2 = 0 \n\]\n\n\[ \n{5}^{{3x} - 1} = 2 \n\]\n\n\[ \n{\log }_{5}\left( {5}^{{3x} - 1}\right) = {\log }_{5}\left( 2\right) \n\]\n\n... | Yes |
Solve \( {4}^{{2y} + 1} = {2}^{y - 1} \) for \( y \) . | With these types of problems, we want to look at both bases and see if they are related in any way. Here, we have a base of 2 on the right and a base of 4 on the left. You’ll probably notice that \( 4 = {2}^{2} \) ; we can use this to our advantage when solving. Let's start by rewriting our statement using this fact.\n... | Yes |
Solve \( {\log }_{5}\left( {{2x} + 3}\right) = 2 \) for \( x \) . | Here, the logarithm is already isolated on one side, so we can start off by using the definition of logarithms shown in equation 1.39 to remove the logarithm from our equation.\n\n\[ \n{\log }_{5}\left( {{2x} + 3}\right) = 2\text{, then, from our definition,} \n\]\n\n\[ \n{2x} + 3 = {5}^{2} \n\]\n\n\[ \n{2x} + 3 = {25}... | Yes |
Write the equation for the line that passes through \( \left( {2,4}\right) \) and is parallel to \( {3x} + y = 6 \) in slope-intercept form. | Solution Let's analyze the information we so far. First, we know that our line goes through the point \( \left( {2,4}\right) \) . We know we also need its slope.\n\nWe’re told it’s parallel to \( {3x} + y = 6 \) , so it will have the same slope as that line. However, this line isn't in either of our forms, so it's not ... | Yes |
Write the equation for the line that passes through \( \\left( {2,4}\\right) \) and is parallel to \( {3x} + y = 6 \) in slope-intercept form. | Solution From our previous work on this problem, we know that our line goes through the point \( \\left( {2,4}\\right) \) and that it has a slope of -3 . We also know that slope-intercept form looks like\n\n\[ y = {mx} + b \]\n\nand that this statement holds for every \( \\left( {x, y}\\right) \) pair that are on the l... | Yes |
In slope-intercept form, write the equation of a line with a y-intercept of 5 that is perpendicular to \( y - 4 = 6\left( {x - 2}\right) \) . | Solution Let's look at the information we have so far. We are told that the line we are interested in has a y-intercept of 5 . We know that \( x = 0 \) for the y-intercept, so this really means that the y-intercept is the point \( \left( {0,5}\right) \) . Next, we know that we are perpendicular to the line \( y - 4 = 6... | Yes |
Determine if each of the following numbers is included in the interval \( \lbrack - 5,{27}) \) . | Solution For each of these, we need to determine if the number is between the two numbers given in the interval.\n\n1. 2 is bigger than -5 and smaller than 27 so it is in the interval.\n\n2. \( \pi \) is bigger than -5 and smaller than 27 so it is in the interval.\n\n3. -5 is one of our endpoints, so we need to see if ... | Yes |
Solve \( {x}^{2} - {6x} + 8 > 0 \) . | Solution Our first step is to convert this into an equality statement by changing the \( > \) symbol to an \( = \) symbol:\n\n\[ \n{x}^{2} - {6x} + 8 = 0 \n\]\n\nNow, we can use any solution method we learned for finding the roots of a quadratic function to solve. Here we have a quadratic that factors nicely, so we wil... | Yes |
Determine the domain for each of the following power functions: 1. \( f\left( x\right) = {x}^{2/3} \) 5. \( w\left( t\right) = \frac{1}{2}{t}^{-1/3} \) | Solution For each of these, we need to look at the exponent only;\n\nscalar multiplication of a function does not affect the domain.\n\n1. For \( f\left( x\right), b = \frac{2}{3} \) . This is a fraction, so we need to look at the denominator. The denominator is 3 , an odd number. This tells us that negative inputs are... | No |
Determine the domain of \[ f\left( \theta \right) = \frac{{\theta }^{2} + 4}{\sqrt{\theta } - 1} \] | Solution First, let's look at each of the individual functions. The numerator has \( {\theta }^{2} + 4 \) . This is the addition of two monomials: \( {\theta }^{2} \) and 4 . The addition doesn’t introduce any problems. \( {\theta }^{2} \) is a power function where \( b \) is a positive whole number, so it doesn't intr... | Yes |
Determine the domain of\n\n\[ f\left( x\right) = \sqrt{6 - x} + {12x} \] | Solution Overall, we have the addition of two functions, \( \sqrt{6 - x} \) and \( {12x} \) . Addition doesn’t introduce any problems. The second function, \( {12x} \) has no problem inputs because it’s a power function where \( b \) is a positive whole number. However, we know that the square root function can't use n... | Yes |
Graph the line \( y = {2x} - 3 \) | Solution Let's start by identifying the slope and a point. We are in slope-intercept form, so we can see that the slope is \( m = 2 \) and the y-intercept \( \mid \) is \( \left( {0, - 3}\right) \) . We’ll starting by plotting a point at \( \left( {0, - 3}\right) \) . Then, we’ll move to the right 1 unit and up 2 units... | Yes |
Graph the function \( f\left( x\right) = {x}^{2} + 3 \) . | Here we can see that we have a function with a constant added to it; this tells us that we need to graph \( {x}^{2} \), but with a vertical shift. The constant, +3 tells us that we will take this base function and shift it up 3 units (if this was -3 we would shift the function down by 3 units). \n\n = 2\sin \left( x\right) - 1 \) . | Solution First, let's identify our base function. Here, we are working with \( \sin \left( x\right) \) . We see that we are multiplying by 2 and subtracting 1 ; this tells us we have a vertical stretch and a vertical shift. Which should we do first? The answer comes from our order of operations: multiplication should b... | No |
Graph the function \( f\left( x\right) = 4{x}^{2} + {4x} + 1 = {\left( 2x + 1\right) }^{2} \) . | Solution Our base function here is \( {x}^{2} \) (dotted). We don’t have any vertical modifications, just horizontal modifications since everything is happening inside the function. We see we have a shift left of 1 , due to the +1 , and then we need to shrink by a factor of 2 since \( x \) is multiplied by 2 . First, w... | Yes |
Graph the function \( g\left( x\right) = - {\left( \frac{1}{2}x + 2\right) }^{2} + 3 \) . | Solution As with our previous examples, the first step is to identify the base function. Here our base functions is \( {x}^{2} \) (dotted). We said that we should start with horizontal changes, so let's look at those first. With horizontal modifications, we need to work with the shift and then the stretch. Inside of ou... | Yes |
Determine the equation for the graph of \( f\left( x\right) = {x}^{3} \) after it has been shifted 2 units to the right, flipped vertically, and shifted 2 units up. | Solution Here the transformations have been given in the same order that we would apply them. Our first step, is to shift the function to the right, so we will change to \( {\left( x - 2\right) }^{3} \) . Next, we want to flip the function vertically, so we get \( - {\left( x - 2\right) }^{3} \) . Finally, we want to s... | Yes |
Write \( f\left( x\right) = {x}^{2} + {4x} + 6 \) in the form \( {\left( x + a\right) }^{2} + b \) . | Solution We saw in equation 2.6 that \( {\left( x + a\right) }^{2} = {x}^{2} + {2ax} + {a}^{2} \) . We’ll use the coefficient on \( x \) from our function to determine \( a \) .\n\n\( \ln f\left( x\right), x \) has a coefficient of 4 and in the expanded pattern \( x \) has a coefficient of \( {2a} \) . We want these to... | Yes |
Complete the square for \( g\left( t\right) = {t}^{2} - {7t} + {10} \) . | Solution There is one big difference between this problem and our previous example: our input variable has changed. That means instead of our goal looking like \( {\left( x + a\right) }^{2} + b \), our goal looks like \( {\left( t + a\right) }^{2} + b \) . Regardless, we’ll follow the same thought process we used in th... | Yes |
Write \( f\left( x\right) = 4{x}^{2} + {12x} - 3 \) in the form \( c{\left( x + a\right) }^{2} + b \) . | Solution Since we want our answer in the form \( c{\left( x + a\right) }^{2} + b \), we will use equation 2.8. In equation 2.8, we see that the coefficient on \( {x}^{2} \) in the expanded form is \( c \) . For \( f\left( x\right) \), the \( {x}^{2} \) coefficient is 4, so we have \( c = 4 \) .\n\nNext, we’ll work with... | Yes |
Write \( f\left( x\right) = 4{x}^{2} + {12x} - 3 \) in the form \( c{\left( x + a\right) }^{2} + b \) . | Solution We'll start by factoring 4 out from the equation and completing the square on the remaining quadratic factor. By factoring out the \( 4,{x}^{2} \) will have a coefficient of 1 and we can work like we did in our earlier examples.\n\n\[ f\left( x\right) = 4{x}^{2} + {12x} - 3 = 4\left\lbrack {{x}^{2} + {3x} - \f... | Yes |
The statement \( {y}^{3}\sin \left( w\right) = 4{y}^{2}z + 2{x}^{2}y + {xy} + {10} \) is what type of statement in terms of\n\n1. the letter \( w \) ? 3. the letter \( y \) ?\n\n2. the letter \( x \) ? 4. the letter \( z \) ? | 1. Here, our focus is on the letter \( w \) . The letter \( w \) only appears on the left-hand side in the term \( {y}^{3}\sin \left( w\right) \) . Since \( w \) is inside of the sine function,\n\nThis statement is trigonometric in \( w \)\n\n2. Here, our focus is on the letter \( x \) . The letter \( x \) appears in t... | Yes |
Solve the statement \( {xz} + {2yz} - 4 = \sin \left( x\right) + {y}^{2} + {y}^{2}z \) for \( z \) . | Here we see that we do have a linear statement in \( z \) : the highest degree of \( z \) in the statement is 1, and \( z \) does not appear inside of any other functions. We’ll start by gathering every term with a \( z \) on the left side by subtracting \( {y}^{2}z \) from both sides:\n\n\[ {xz} + {2yz} + {y}^{2}z - 4... | Yes |
Solve \( x\left( {x - 4}\right) = - {13} \) for \( x \) . | Solution From the form of the statement we are given, it's not entirely obvious that we are working with a quadratic. We'll need to expand the left side before we do anything else so that we can easily see that it is a quadratic and so we can correctly determine all of the coefficients. Expanding gives \( {x}^{2} - {4x... | Yes |
Find all points of intersection of \( x - 4 = {y}^{2} \) and \( {x}^{2} - {4x} = - {y}^{2} \) . | Solution Here we can see that the only \ | No |
Find all points of intersection of \( f\left( x\right) = {x}^{2} + 1 \) and \( g\left( x\right) = x + 1 \) | Solution Here, both equations are given using function notation; this means that really \( f\left( x\right) \) tells us the value of the \( y \) coordinate at \( x \), so we can replace it with \( y : y = {x}^{2} + 1 \) . Similarly, \( g\left( x\right) \) tells us the value of the \( y \) coordinate at \( x \) for the ... | Yes |
Find all points of intersection of \( {2x} + {3y} = 2 \) and \( - x + y = 4 \) . | Solution We will first try to eliminate \( x \) from both equations. The first equation has \( {2x} \) and the second has \( - x \) . If we multiply the second equation by 2 and add it to the first, the \( x \) terms will cancel out:\n\n\[ \n{2x} + {3y} = 2 \n\]\n\n\[ \n+ 2\left( {-x + y = 4}\right) \n\]\n\nor:\n\n\[ \... | Yes |
Find all points of intersection of \( {2x} + {3y} = 2 \) and \( - x + y = 4 \) . | The first equation has \( {3y} \) and the second has \( y \) . We will multiply the first equation by \( - \frac{1}{3} \) and add it to the second equation:\n\n\[ \n- \frac{1}{3}\left( {{2x} + {3y} = 2}\right) \n\]\n\n\[ \n\begin{array}{r} + \;\left( {-x + y = 4}\right) \\ - \frac{5}{3}x = \frac{10}{3} \end{array} \n\]... | Yes |
Simplify \( \frac{3}{x + 2} - \frac{x + 1}{x - 2} \) . | Solution First, we will multiply by the missing factors. We will multiply the first term by \( \frac{x - 2}{x - 2} \) and the second term by \( \frac{x + 2}{x + 2} \) . This gives us:\n\n\[ \frac{3}{x + 2} - \frac{x + 1}{x - 2} = \frac{3}{x + 2} \times \frac{x - 2}{x - 2} - \frac{x + 1}{x - 2} \times \frac{x + 2}{x + 2... | Yes |
Simplify \( \frac{2{x}^{3} + {10}{x}^{2} + {12x}}{2{x}^{3} - {8x}} \) . | Solution The first step here is to factor both the numerator and the denominator. We won't show those steps here, but you should verify our result. Once we have factored both, we will see if we have any common factors; if we do we can remove them from both the numerator and the denominator.\n\n\[ \frac{2{x}^{3} + {10}{... | Yes |
Simplify \( \frac{x + 1}{\frac{x - 1}{{x}^{2}}} \) . | \[ \frac{x + 1}{\frac{x - 1}{{x}^{2}}} = \left( {x + 1}\right) \div \frac{x - 1}{{x}^{2}} \]\n\[ = \left( {x + 1}\right) \times \frac{{x}^{2}}{x - 1} \]\n\[ = \frac{\left( {x + 1}\right) \left( {x}^{2}\right) }{x - 1} \]\n\[ = \frac{{x}^{3} + {x}^{2}}{x - 1} \]\nThere are no common factors, so we are done, and our fina... | Yes |
Simplify \( \frac{\frac{2}{x + 1}}{x + 2} \) . | Solution Here, the nested fraction is in the numerator. For this case, we can simply rewrite a little bit; instead of dividing by \( x + 2 \), we can multiply by \( \frac{1}{x + 2} \) . This is analogous to multiplying by one half instead of dividing by 2 ; both have the same meaning.\n\n\[ \frac{\frac{2}{x + 1}}{x + 2... | Yes |
Simplify \( \frac{\frac{x}{x + 1}}{\frac{x - 2}{x - 1}} \) . | Solution Here we will use the ideas from both of the previous examples. We will multiply the numerator by the reciprocal of the denominator:\n\n\[ \frac{\frac{x}{x + 1}}{\frac{x - 2}{x - 1}} = \frac{x}{x + 1} \div \frac{x - 2}{x - 1} \]\n\n\[ = \frac{x}{x + 1} \times \frac{x - 1}{x - 2} \]\n\n\[ = \frac{\left( x\right)... | Yes |
Suppose that \( \cos \left( \theta \right) = \frac{12}{13} \) . Determine all possible values of \( \sin \left( \theta \right) \) . | To help find the possible values of \( \sin \left( \theta \right) \), we will draw a right triangle and label it using the values we already know. We know that \( \cos \left( \theta \right) = \frac{12}{13} \), so we can use 12 as the length of the adjacent side and 13 as the length of the hypotenuse:\n\n\n\nOur next step is to change the entry in the box to a 1 . To do this, let's multiply row \( 1 \) by \( - \frac{1}{3} \) .\n\n\[ \left\lbrac... | Yes |
Example 5 Use Gaussian elimination to put the matrix \( A \) into reduced row echelon form, where\n\n\[ A = \left\lbrack \begin{matrix} - 2 & - 4 & - 2 & - {10} & 0 \\ 2 & 4 & 1 & 9 & - 2 \\ 3 & 6 & 1 & {13} & - 4 \end{matrix}\right\rbrack \] | Solution We start by wanting to make the entry in the first column and first row a 1 (a leading 1). To do this we’ll scale the first row by a factor of \( - \frac{1}{2} \) .\n\n\[ - \frac{1}{2}{R}_{1} \rightarrow {R}_{1}\;\left\lbrack \begin{matrix} 1 & 2 & 1 & 5 & 0 \\ 2 & 4 & 1 & 9 & - 2 \\ 3 & 6 & 1 & {13} & - 4 \en... | Yes |
Example 6 Put the matrix\n\n\\[ \n\\left\\lbrack \\begin{array}{llll} 1 & 2 & 1 & 3 \\\\ 2 & 1 & 1 & 1 \\\\ 3 & 3 & 2 & 1 \\end{array}\\right\\rbrack \n\\]\n\ninto reduced row echelon form. | Solution Here we will show all steps without explaining each one.\n\n\\[ \n- 2{R}_{1} + {R}_{2} \\rightarrow {R}_{2} \n\\]\n\n\\[ \n\\left\\lbrack \\begin{matrix} 1 & 2 & 1 & 3 \\\\ 0 & - 3 & - 1 & - 5 \\\\ 0 & - 3 & - 1 & - 8 \\end{matrix}\\right\\rbrack \n\\]\n\n\\[ \n- 3{R}_{1} + {R}_{3} \\rightarrow {R}_{3} \n\\]\n... | Yes |
Example 7 Put the matrix \( A \) into reduced row echelon form, where\n\n\[ A = \left\lbrack \begin{matrix} 2 & 1 & - 1 & 4 \\ 1 & - 1 & 2 & {12} \\ 2 & 2 & - 1 & 9 \end{matrix}\right\rbrack \] | Solution We'll again show the steps without explanation, although we will stop at the end of the forward steps and make a comment.\n\n\[ \left\lbrack \begin{matrix} 1 & 1/2 & - 1/2 & 2 \\ 1 & - 1 & 2 & {12} \\ 2 & 2 & - 1 & 9 \end{matrix}\right\rbrack \]\n\n\[ - {R}_{1} + {R}_{2} \rightarrow {R}_{2} \]\n\n\[ \left\lbra... | Yes |
Every linear system of equations has exactly one solution, infinite solutions, or no solution. | How can we tell what kind of solution (if one exists) a given system of linear equations has? The answer to this question lies with properly understanding the reduced row echelon form of a matrix. To discover what the solution is to a linear system, we first put the matrix into reduced row echelon form and then interpr... | "No" |
Find the solution to the linear system\n\n\[ \n{x}_{1} + {x}_{2} = 1 \n\]\n\n\[ \n2{x}_{1} + 2{x}_{2} = 2 \n\] | Solution Create the corresponding augmented matrix, and then put the matrix into reduced row echelon form.\n\n\[ \n\left\lbrack \begin{array}{lll} 1 & 1 & 1 \\ 2 & 2 & 2 \end{array}\right\rbrack \;\overrightarrow{\operatorname{rref}}\;\left\lbrack \begin{array}{lll} 1 & 1 & 1 \\ 0 & 0 & 0 \end{array}\right\rbrack \n\]\... | Yes |
Example 9 Find the solution to the linear system\n\n\[ \n{x}_{2}\; - \;{x}_{3}\; = \;3 \]\n\n\[ \n{x}_{1}\; + \;2{x}_{3}\; = \;2 \]\n\n\[ \n- 3{x}_{2} + 3{x}_{3} = - 9 \]\n | Solution To find the solution, put the corresponding matrix into reduced\n\nrow echelon form.\n\n\[ \n\left\lbrack \begin{matrix} 0 & 1 & - 1 & 3 \\ 1 & 0 & 2 & 2 \\ 0 & - 3 & 3 & - 9 \end{matrix}\right\rbrack \;\overrightarrow{\operatorname{rref}}\;\left\lbrack \begin{matrix} 1 & 0 & 2 & 2 \\ 0 & 1 & - 1 & 3 \\ 0 & 0 ... | Yes |
Find the solution to the linear system\n\n\[ \n{x}_{1} + {x}_{2} + {x}_{3} = 1 \n\]\n\n\[ \n{x}_{1} + 2{x}_{2} + {x}_{3} = 2\text{.} \n\]\n\n\[ \n2{x}_{1} + 3{x}_{2} + 2{x}_{3} = 0 \n\] | Solution We start by putting the corresponding matrix into reduced row\nechelon form.\n\n\[ \n\left\lbrack \begin{array}{llll} 1 & 1 & 1 & 1 \\ 1 & 2 & 1 & 2 \\ 2 & 3 & 2 & 0 \end{array}\right\rbrack \;\overrightarrow{\operatorname{rref}}\;\left\lbrack \begin{array}{llll} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1... | Yes |
Example 11 Confirm that the linear system\n\n\\[ \nx + y = 0 \n\\]\n\n\\[ \n{2x} + {2y} = 4 \n\\]\n\nhas no solution. | Solution We can verify that this system has no solution in two ways. First, let’s just think about it. If \( x + y = 0 \), then it stands to reason, by multiplying both sides of this equation by 2, that \( {2x} + {2y} = 0 \) . However, the second equation of our system says that \( {2x} + {2y} = 4 \) . Since \( 0 \neq ... | Yes |
Give the solution to a linear system whose augmented matrix in reduced row echelon form is\n\n\[ \left\lbrack \begin{matrix} 1 & - 1 & 0 & 2 & 4 \\ 0 & 0 & 1 & - 3 & 7 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}\right\rbrack \]\n\nand give two particular solutions. | Solution We can essentially ignore the third row; it does not divulge any information about the solution. \( {}^{4} \) The first and second rows can be rewritten as the following equations:\n\n\[ {x}_{1} - {x}_{2} + 2{x}_{4} = 4 \]\n\n\[ {x}_{3} - 3{x}_{4} = 7\text{.} \]\n\nNotice how the variables \( {x}_{1} \) and \(... | No |
Find the solution to the linear system\n\n\[ \n{x}_{1} + {x}_{2} + {x}_{3} = 5 \n\]\n\n\[ \n{x}_{1} - {x}_{2} + {x}_{3} = 3 \n\]\n\nand give two particular solutions. | Solution The corresponding augmented matrix and its reduced row echelon form are given below.\n\n\[ \n\left\lbrack \begin{matrix} 1 & 1 & 1 & 5 \\ 1 & - 1 & 1 & 3 \end{matrix}\right\rbrack \;\overrightarrow{\operatorname{rref}}\;\left\lbrack \begin{array}{llll} 1 & 0 & 1 & 4 \\ 0 & 1 & 0 & 1 \end{array}\right\rbrack \n... | Yes |
For what values of \( k \) will the given system have exactly one solution, infinite solutions, or no solution? | Solution We answer this question by forming the augmented matrix and starting the process of putting it into reduced row echelon form. Below we see the augmented matrix and one elementary row operation that starts the Gaussian elimination process.\n\n\[ \left\lbrack \begin{array}{lll} 1 & 2 & 3 \\ 3 & k & 9 \end{array}... | Yes |
Example 16 A jar contains 100 blue, green, red and yellow marbles. There are twice as many yellow marbles as blue; there are 10 more blue marbles than red; the sum of the red and yellow marbles is the same as the sum of the blue and green. How many marbles of each color are there? | Solution Let’s call the number of blue balls \( b \), and the number of the other balls \( g, r \) and \( y \), each representing the obvious. Since we know that we have 100 marbles, we have the equation\n\n\[ b + g + r + y = {100}. \]\n\nThe next sentence in our problem statement allows us to create three more equatio... | No |
Example 17 A concert hall has seating arranged in three sections. As part of a special promotion, guests will recieve two of three prizes. Guests seated in the first and second sections will receive Prize A, guests seated in the second and third sections will receive Prize B, and guests seated in the first and third se... | Solution Before we rush in and start making equations, we should be clear about what is being asked. The final sentence asks: \ | No |
Find the equation of the quadratic function that goes through the points \( \left( {-1,6}\right) ,\left( {1,2}\right) \) and \( \left( {2,3}\right) \) . | This may not seem like a \ | No |
A woman has \( {32}\$ 1,\$ 5 \) and \( \$ {10} \) bills in her purse, giving her a total of \( \$ {100} \) . How many bills of each denomination does she have? | Solution Let’s name our unknowns \( x, y \) and \( z \) for our ones, fives and tens, respectively (it is tempting to call them \( o, f \) and \( t \), but \( o \) looks too much like 0 ). We know that there are a total of 32 bills, so we have the equation\n\n\[ x + y + z = {32}. \]\n\nWe also know that we have \( \$ {... | No |
In a football game, teams can score points through touchdowns worth 6 points, extra points (that follow touchdowns) worth 1 point, two point conversions (that also follow touchdowns) worth 2 points and field goals, worth 3 points. You are told that in a football game, the two competing teams scored on 7 occasions, givi... | Solution The question asks how the points were scored; we can interpret this as asking how many touchdowns, extra points, two point conversions and field goals were scored. We’ll need to assign variable names to our unknowns; let \( t \) represent the number of touchdowns scored; let \( x \) represent the number of ext... | No |
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