Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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Simplify: | In summary, adding and subtracting complex numbers results in a complex number. | No |
Multiply: | Solution: | No |
1. \( \{ x \mid x \leq - 2 \) or \( x \geq 2\} \) | The best way to proceed here is to graph the set of numbers on the number line and glean the answer from it. The inequality \( x \leq - 2 \) corresponds to the interval \( ( - \infty , - 2\rbrack \) and the inequality \( x \geq 2 \) corresponds to the interval \( \lbrack 2,\infty ) \) . Since we are looking to describe... | Yes |
Plot the following points: \( A\\left( {5,8}\\right), B\\left( {-\\frac{5}{2},3}\\right), C\\left( {-{5.8}, - 3}\\right), D\\left( {{4.5}, - 1}\\right), E\\left( {5,0}\\right) \) , \( F\\left( {0,5}\\right), G\\left( {-7,0}\\right), H\\left( {0, - 9}\\right), O\\left( {0,0}\\right) .{}^{10} \) | Solution. To plot these points, we start at the origin and move to the right if the \( x \) -coordinate is positive; to the left if it is negative. Next, we move up if the \( y \) -coordinate is positive or down if it is negative. If the \( x \) -coordinate is 0, we start at the origin and move along the \( y \) -axis ... | Yes |
Example 1.1.3. Let \( P \) be the point \( \left( {-2,3}\right) \) . Find the points which are symmetric to \( P \) about the:\n\n1. \( x \) -axis 2. \( y \) -axis 3. origin\n\nCheck your answer by plotting the points. | Solution. The figure after Definition 1.3 gives us a good way to think about finding symmetric points in terms of taking the opposites of the \( x \) - and/or \( y \) -coordinates of \( P\left( {-2,3}\right) \) .\n\n---\n\n1. To find the point symmetric about the \( x \) -axis, we replace the \( y \) -coordinate with i... | Yes |
Find and simplify the distance between \( P\left( {-2,3}\right) \) and \( Q\left( {1, - 3}\right) \) . | \[ d = \sqrt{{\left( {x}_{1} - {x}_{0}\right) }^{2} + {\left( {y}_{1} - {y}_{0}\right) }^{2}} \] \[ = \sqrt{{\left( 1 - \left( -2\right) \right) }^{2} + {\left( -3 - 3\right) }^{2}} \] \[ = \sqrt{9 + {36}} \] \[ = 3\sqrt{5} \] So the distance is \( 3\sqrt{5} \) . | Yes |
Find all of the points with \( x \) -coordinate 1 which are 4 units from the point \( \\left( {3,2}\\right) \) . | Solution. We shall soon see that the points we wish to find are on the line \( x = 1 \), but for now we’ll just view them as points of the form \( \\left( {1, y}\\right) \). Visually,\n\n\n\nWe require that the distanc... | Yes |
Find the midpoint of the line segment connecting \( P\left( {-2,3}\right) \) and \( Q\left( {1, - 3}\right) \) . | \[ M = \left( {\frac{{x}_{0} + {x}_{1}}{2},\frac{{y}_{0} + {y}_{1}}{2}}\right) \] \[ = \left( {\frac{\left( {-2}\right) + 1}{2},\frac{3 + \left( {-3}\right) }{2}}\right) = \left( {-\frac{1}{2},\frac{0}{2}}\right) \] \[ = \left( {-\frac{1}{2},0}\right) \] The midpoint is \( \left( {-\frac{1}{2},0}\right) \) . | Yes |
Example 1.1.7. If \( a \neq b \), prove that the line \( y = x \) equally divides the line segment with endpoints \( \left( {a, b}\right) \) and \( \left( {b, a}\right) \) . | Solution. To prove the claim, we use Equation 1.2 to find the midpoint\n\n\[ M = \left( {\frac{a + b}{2},\frac{b + a}{2}}\right) \]\n\n\[ = \left( {\frac{a + b}{2},\frac{a + b}{2}}\right) \]\n\nSince the \( x \) and \( y \) coordinates of this point are the same, we find that the midpoint lies on the line \( y = x \), ... | Yes |
1. \( A = \{ \left( {0,0}\right) ,\left( {-3,1}\right) ,\left( {4,2}\right) ,\left( {-3,2}\right) \} \) | 1. To graph \( A \), we simply plot all of the points which belong to \( A \), as shown below on the left.\n\n\n\nThe graph of \( A \) | No |
Determine whether or not \( \left( {2, - 1}\right) \) is on the graph of \( {x}^{2} + {y}^{3} = 1 \) . | We substitute \( x = 2 \) and \( y = - 1 \) into the equation to see if the equation is satisfied.\n\n\[{\left( 2\right) }^{2} + {\left( -1\right) }^{3}\overset{?}{ = }1\]\n\n\[3 \neq 1\]\n\nHence, \( \left( {2, - 1}\right) \) is not on the graph of \( {x}^{2} + {y}^{3} = 1 \) . | Yes |
Graph \( {x}^{2} + {y}^{3} = 1 \) . | To efficiently generate points on the graph of this equation, we first solve for \( y \)\n\n\[ \n{x}^{2} + {y}^{3} = 1 \n\]\n\n\[ \n{y}^{3} = 1 - {x}^{2} \n\]\n\n\[ \n\sqrt[3]{{y}^{3}} = \sqrt[3]{1 - {x}^{2}} \n\]\n\n\[ \ny = \sqrt[3]{1 - {x}^{2}} \n\]\n\nWe now substitute a value in for \( x \), determine the correspo... | Yes |
Find the \( x \) - and \( y \) -intercepts (if any) of the graph of \( {\left( x - 2\right) }^{2} + {y}^{2} = 1 \) . Test for symmetry. Plot additional points as needed to complete the graph. | Solution. To look for \( x \) -intercepts, we set \( y = 0 \) and solve\n\n\[ \n{\left( x - 2\right) }^{2} + {y}^{2} = 1 \n\] \n\n\[ \n{\left( x - 2\right) }^{2} + {0}^{2} = 1 \n\] \n\n\[ \n{\left( x - 2\right) }^{2} = 1 \n\] \n\n\[ \n\sqrt{{\left( x - 2\right) }^{2}} = \sqrt{1}\;\text{extract square roots} \n\] \n\n\[... | Yes |
Example 1.3.1. Which of the following relations describe \( y \) as a function of \( x \) ?\n\n1. \( {R}_{1} = \{ \left( {-2,1}\right) ,\left( {1,3}\right) ,\left( {1,4}\right) ,\left( {3, - 1}\right) \} \) 2. \( {R}_{2} = \{ \left( {-2,1}\right) ,\left( {1,3}\right) ,\left( {2,3}\right) ,\left( {3, - 1}\right) \} \) | Solution. A quick scan of the points in \( {R}_{1} \) reveals that the \( x \) -coordinate 1 is matched with two different \( y \) -coordinates: namely 3 and 4. Hence in \( {R}_{1}, y \) is not a function of \( x \) . On the other hand, every \( x \) -coordinate in \( {R}_{2} \) occurs only once which means each \( x \... | Yes |
Use the Vertical Line Test to determine which of the following relations describes \( y \) as a function of \( x \) . | Solution. Looking at the graph of \( R \), we can easily imagine a vertical line crossing the graph more than once. Hence, \( R \) does not represent \( y \) as a function of \( x \) . However, in the graph of \( S \), every vertical line crosses the graph at most once, so \( S \) does represent \( y \) as a function o... | Yes |
Use the Vertical Line Test to determine which of the following relations describes \( y \) as a function of \( x \) . | Both \( {S}_{1} \) and \( {S}_{2} \) are slight modifications to the relation \( S \) in the previous example whose graph we determined passed the Vertical Line Test. In both \( {S}_{1} \) and \( {S}_{2} \), it is the addition of the point \( \left( {1,2}\right) \) which threatens to cause trouble. In \( {S}_{1} \), th... | Yes |
Find the domain and range of the function \( F = \{ \left( {-3,2}\right) ,\left( {0,1}\right) ,\left( {4,2}\right) ,\left( {5,2}\right) \} \) and of the function \( G \) whose graph is given above on the right. | Solution. The domain of \( F \) is the set of the \( x \) -coordinates of the points in \( F \), namely \( \{ - 3,0,4,5\} \) and the range of \( F \) is the set of the \( y \) -coordinates, namely \( \{ 1,2\} \) . To determine the domain and range of \( G \), we need to determine which \( x \) and \( y \) values occur ... | Yes |
Example 1.3.5. Determine which equations represent \( y \) as a function of \( x \) . | Solution. For each of these equations, we solve for \( y \) and determine whether each choice of \( x \) will determine only one corresponding value of \( y \) .\n\n1.\n\n\[ \n{x}^{3} + {y}^{2} = 1 \n\] \n\n\[ \n{y}^{2} = 1 - {x}^{3} \n\] \n\n\[ \n\sqrt{{y}^{2}} = \sqrt{1 - {x}^{3}}\;\text{extract square roots} \n\] \n... | Yes |
Suppose a function \( g \) is described by applying the following steps, in sequence\n\n1. add 4\n\n2. multiply by 3\n\nDetermine \( g\left( 5\right) \) and find an expression for \( g\left( x\right) \) . | Solution. Starting with 5, step 1 gives \( 5 + 4 = 9 \) . Continuing with step 2, we get \( \left( 3\right) \left( 9\right) = {27} \) . To find a formula for \( g\left( x\right) \), we start with our input \( x \) . Step 1 produces \( x + 4 \) . We now wish to multiply this entire quantity by 3, so we use a parentheses... | Yes |
Example 1.4.2. Let \( f\left( x\right) = - {x}^{2} + {3x} + 4 \)\n\n1. Find and simplify the following.\n\n(a) \( f\left( {-1}\right), f\left( 0\right), f\left( 2\right) \)\n\n(b) \( f\left( {2x}\right) ,{2f}\left( x\right) \)\n\n(c) \( f\left( {x + 2}\right), f\left( x\right) + 2, f\left( x\right) + f\left( 2\right) \... | Solution.\n\n1. (a) To find \( f\left( {-1}\right) \), we replace every occurrence of \( x \) in the expression \( f\left( x\right) \) with -1\n\n\[ f\left( {-1}\right) = - {\left( -1\right) }^{2} + 3\left( {-1}\right) + 4 \]\n\n\[ = - \left( 1\right) + \left( {-3}\right) + 4 \]\n\n\[ = 0 \]\n\nSimilarly, \( f\left( 0\... | Yes |
Example 1.4.3. Find the domain \( {}^{3} \) of the following functions.\n\n1. \( g\left( x\right) = \sqrt{4 - {3x}} \) 2. \( h\left( x\right) = \sqrt[5]{4 - {3x}} \) | 1. The potential disaster for \( g \) is if the radicand \( {}^{4} \) is negative. To avoid this, we set \( 4 - {3x} \geq 0 \) . From this, we get \( {3x} \leq 4 \) or \( x \leq \frac{4}{3} \) . What this shows is that as long as \( x \leq \frac{4}{3} \), the expression \( 4 - {3x} \geq 0 \), and the formula \( g\left(... | Yes |
1. Find and interpret \( h\left( {10}\right) \) and \( h\left( {60}\right) \) . | We first note that the independent variable here is \( t \), chosen because it represents time. Secondly, the function is broken up into two rules: one formula for values of \( t \) between 0 and 20 inclusive, and another for values of \( t \) greater than 20 . Since \( t = {10} \) satisfies the inequality \( 0 \leq t ... | Yes |
1. Find \( \left( {f + g}\right) \left( {-1}\right) \) | To find \( \left( {f + g}\right) \left( {-1}\right) \) we first find \( f\left( {-1}\right) = 8 \) and \( g\left( {-1}\right) = 4 \) . By definition, we have that \( \left( {f + g}\right) \left( {-1}\right) = f\left( {-1}\right) + g\left( {-1}\right) = 8 + 4 = {12}. \) | Yes |
Find and simplify the difference quotients for the following functions 1. \( f\left( x\right) = {x}^{2} - x - 2 \) | To find \( f\left( {x + h}\right) \), we replace every occurrence of \( x \) in the formula \( f\left( x\right) = {x}^{2} - x - 2 \) with the quantity \( \left( {x + h}\right) \) to get \[ f\left( {x + h}\right) = {\left( x + h\right) }^{2} - \left( {x + h}\right) - 2 \] \[ = {x}^{2} + {2xh} + {h}^{2} - x - h - 2\text{... | Yes |
1. Find and interpret \( C\left( 0\right) \) . | We substitute \( x = 0 \) into the formula for \( C\left( x\right) \) and get \( C\left( 0\right) = {100}\left( 0\right) + {2000} = {2000} \) . This means to produce 0 dOpis, it costs \$2000. In other words, the fixed (or start-up) costs are \$2000. The reader is encouraged to contemplate what sorts of expenses these m... | Yes |
Example 1.6.1. Graph \( f\left( x\right) = {x}^{2} - x - 6 \) . | Solution. To graph \( f \), we graph the equation \( y = f\left( x\right) \) . To this end, we use the techniques outlined in Section 1.2.1. Specifically, we check for intercepts, test for symmetry, and plot additional points as needed. To find the \( x \) -intercepts, we set \( y = 0 \) . Since \( y = f\left( x\right)... | Yes |
Graph: \( f\left( x\right) = \left\{ \begin{matrix} 4 - {x}^{2} & \text{ if } & x < 1 \\ x - 3, & \text{ if } & x \geq 1 \end{matrix}\right. \) | Solution. We proceed as before - finding intercepts, testing for symmetry and then plotting additional points as needed. To find the \( x \) -intercepts, as before, we set \( f\left( x\right) = 0 \) . The twist is that we have two formulas for \( f\left( x\right) \) . For \( x < 1 \), we use the formula \( f\left( x\ri... | Yes |
Example 1.6.3. Determine analytically if the following functions are even, odd, or neither even nor odd. Verify your result with a graphing calculator. | 1.\n\n\[ f\left( x\right) = \frac{5}{2 - {x}^{2}} \]\n\n\[ f\left( {-x}\right) = \frac{5}{2 - {\left( -x\right) }^{2}} \]\n\n\[ f\left( {-x}\right) = \frac{5}{2 - {x}^{2}} \]\n\n\[ f\left( {-x}\right) = f\left( x\right) \]\n\nHence, \( f \) is even. The graphing calculator furnishes the following.\n\n \n\nUsing the Minimum and Maximum features, we ge... | Yes |
Example 1.6.6. Find the points on the graph of \( y = {\left( x - 3\right) }^{2} \) which are closest to the origin. Round your answers to two decimal places. | Solution. Suppose a point \( \left( {x, y}\right) \) is on the graph of \( y = {\left( x - 3\right) }^{2} \) . Its distance to the origin \( \left( {0,0}\right) \) is given by\n\n\[ d = \sqrt{{\left( x - 0\right) }^{2} + {\left( y - 0\right) }^{2}} \]\n\n\[ = \sqrt{{x}^{2} + {y}^{2}} \]\n\n\[ = \sqrt{{x}^{2} + {\left\l... | Yes |
Theorem 1.2. Vertical Shifts. Suppose \( f \) is a function and \( k \) is a positive number. | - To graph \( y = f\\left( x\\right) + k \), shift the graph of \( y = f\\left( x\\right) \) up \( k \) units by adding \( k \) to the \( y \) -coordinates of the points on the graph of \( f \) .\n\n- To graph \( y = f\\left( x\\right) - k \), shift the graph of \( y = f\\left( x\\right) \) down \( k \) units by subtra... | Yes |
Theorem 1.3. Horizontal Shifts. Suppose \( f \) is a function and \( h \) is a positive number. | - To graph \( y = f\\left( {x + h}\\right) \), shift the graph of \( y = f\\left( x\\right) \) left \( h \) units by subtracting \( h \) from the \( x \) -coordinates of the points on the graph of \( f \).\n\n- To graph \( y = f\\left( {x - h}\\right) \), shift the graph of \( y = f\\left( x\\right) \) right \( h \) un... | Yes |
1. Graph \( f\left( x\right) = \sqrt{x} \) . Plot at least three points. | Owing to the square root, the domain of \( f \) is \( x \geq 0 \), or \( \lbrack 0,\infty ) \) . We choose perfect squares to build our table and graph below. From the graph we verify the domain of \( f \) is \( \lbrack 0,\infty ) \) and the range of \( f \) is also \( \lbrack 0,\infty ) \) .\n\n<table><thead><tr><th>\... | Yes |
Theorem 1.4. Reflections. Suppose \( f \) is a function. | - To graph \( y = - f\left( x\right) \), reflect the graph of \( y = f\left( x\right) \) across the \( x \) -axis by multiplying the \( y \) -coordinates of the points on the graph of \( f \) by -1 .\n\n- To graph \( y = f\left( {-x}\right) \), reflect the graph of \( y = f\left( x\right) \) across the \( y \) -axis by... | Yes |
Example 1.7.2. Let \( f\left( x\right) = \sqrt{x} \) . Use the graph of \( f \) from Example 1.7.1 to graph the following functions. Also, state their domains and ranges.\n\n1. \( g\left( x\right) = \sqrt{-x} \) | ## Solution.\n\n1. The mere sight of \( \sqrt{-x} \) usually causes alarm, if not panic. When we discussed domains in Section 1.4, we clearly banished negatives from the radicands of even roots. However, we must remember that \( x \) is a variable, and as such, the quantity \( - x \) isn’t always negative. For example,... | Yes |
Theorem 1.5. Vertical Scalings. Suppose \( f \) is a function and \( a > 0 \) . To graph \( y = {af}\left( x\right) \) , multiply all of the \( y \) -coordinates of the points on the graph of \( f \) by \( a \) . We say the graph of \( f \) has been vertically scaled by a factor of \( a \) . | - If \( a > 1 \), we say the graph of \( f \) has undergone a vertical stretching (expansion, dilation) by a factor of \( a \) .\n\n- If \( 0 < a < 1 \), we say the graph of \( f \) has undergone a vertical shrinking (compression, contraction) by a factor of \( \frac{1}{a} \) . | Yes |
Example 1.7.3. Let \( f\left( x\right) = \sqrt{x} \) . Use the graph of \( f \) from Example 1.7.1 to graph the following functions. Also, state their domains and ranges.\n\n1. \( g\left( x\right) = 3\sqrt{x} \) | Solution.\n\n1. First we note that the domain of \( g \) is \( \lbrack 0,\infty ) \) for the usual reason. Next, we have \( g\left( x\right) = {3f}\left( x\right) \) so by Theorem 1.5, we obtain the graph of \( g \) by multiplying all of the \( y \) -coordinates of the points on the graph of \( f \) by 3 . The result i... | Yes |
Theorem 1.7. Transformations. Suppose \( f \) is a function. If \( A \neq 0 \) and \( B \neq 0 \), then to graph\n\n\[ g\left( x\right) = {Af}\left( {{Bx} + H}\right) + K \] | 1. Subtract \( H \) from each of the \( x \) -coordinates of the points on the graph of \( f \) . This results in a horizontal shift to the left if \( H > 0 \) or right if \( H < 0 \) .\n\n2. Divide the \( x \) -coordinates of the points on the graph obtained in Step 1 by \( B \) . This results in a horizontal scaling,... | Yes |
Example 1.7.5. Let \( f\left( x\right) = {x}^{2} \) . Find and simplify the formula of the function \( g\left( x\right) \) whose graph is the result of \( f \) undergoing the following sequence of transformations.\n\n1. Vertical shift up 2 units\n\n2. Reflection across the \( x \) -axis\n\n3. Horizontal shift right 1 u... | Solution. We build up to a formula for \( g\left( x\right) \) using intermediate functions as we’ve seen in previous examples. We let \( {g}_{1} \) take care of our first step. Theorem 1.2 tells us \( {g}_{1}\left( x\right) = f\left( x\right) + 2 = {x}^{2} + 2 \) . Next, we reflect the graph of \( {g}_{1} \) about the ... | Yes |
Find the slope of the line containing the following pairs of points, if it exists. Plot each pair of points and the line containing them. | Solution. In each of these examples, we apply the slope formula, Equation 2.1.\n\n1. \( m = \frac{4 - 0}{2 - 0} = \frac{4}{2} = 2 \) | No |
1. Find the slope of the line containing the points \( \left( {6,{24}}\right) \) and \( \left( {{10},{32}}\right) \) . | 1. For the slope, we have \( m = \frac{{32} - {24}}{{10} - 6} = \frac{8}{4} = 2 \) . | Yes |
Write the equation of the line containing the points \( \\left( {-1,3}\\right) \) and \( \\left( {2,1}\\right) \) . | Solution. In order to use Equation 2.2 we need to find the slope of the line in question so we use Equation 2.1 to get \( m = \\frac{\\Delta y}{\\Delta x} = \\frac{1 - 3}{2 - \\left( {-1}\\right) } = - \\frac{2}{3} \) . We are spoiled for choice for a point \( \\left( {{x}_{0},{y}_{0}}\\right) \) . We’ll use \( \\left(... | Yes |
Graph the following functions. Identify the slope and \( y \) -intercept. | 1. To graph \( f\left( x\right) = 3 \), we graph \( y = 3 \). This is a horizontal line \( \left( {m = 0}\right) \) through \( \left( {0,3}\right) \). | No |
Find and interpret \( C\left( {10}\right) \) . | To find \( C\left( {10}\right) \), we replace every occurrence of \( x \) with 10 in the formula for \( C\left( x\right) \) to get \( C\left( {10}\right) = {80}\left( {10}\right) + {150} = {950} \) . Since \( x \) represents the number of PortaBoys produced, and \( C\left( x\right) \) represents the cost, in dollars, \... | Yes |
1. Find a linear function which fits this data. Use the weekly sales \( x \) as the independent variable and the price \( p \) as the dependent variable. | 1. We recall from Section 1.4 the meaning of ’independent’ and ’dependent’ variable. Since \( x \) is to be the independent variable, and \( p \) the dependent variable, we treat \( x \) as the input variable and \( p \) as the output variable. Hence, we are looking for a function of the form \( p\left( x\right) = {mx}... | Yes |
1. Find and simplify an expression for the weekly revenue \( R\left( x\right) \) as a function of weekly sales \( x \) . | 1. Since \( R = {xp} \), we substitute \( p\left( x\right) = - {1.5x} + {250} \) from Example 2.1.6 to get \( R\left( x\right) = x( - {1.5x} + \) \( {250}) = - {1.5}{x}^{2} + {250x} \) . Since we determined the price-demand function \( p\left( x\right) \) is restricted to \( 0 \leq x \leq {166}, R\left( x\right) \) is ... | Yes |
Theorem 2.1. Properties of Absolute Value: Let \( a, b \) and \( x \) be real numbers and let \( n \) be an integer. \( {}^{a} \) Then\n\n- Product Rule: \( \left| {ab}\right| = \left| a\right| \left| b\right| \) | The proofs of the Product and Quotient Rules in Theorem 2.1 boil down to checking four cases: when both \( a \) and \( b \) are positive; when they are both negative; when one is positive and the other is negative; and when one or both are zero.\n\nFor example, suppose we wish to show that \( \left| {ab}\right| = \left... | Yes |
1. \( \left| {{3x} - 1}\right| = 6 \) | The equation \( \left| {{3x} - 1}\right| = 6 \) is of the form \( \left| x\right| = c \) for \( c > 0 \), so by the Equality Properties, \( \left| {{3x} - 1}\right| = 6 \) is equivalent to \( {3x} - 1 = 6 \) or \( {3x} - 1 = - 6 \) . Solving the former, we arrive at \( x = \frac{7}{3} \) , and solving the latter, we ge... | Yes |
Graph each of the following functions.\n\n1. \( f\left( x\right) = \left| x\right| \) 2. \( g\left( x\right) = \left| {x - 3}\right| \) 3. \( h\left( x\right) = \left| x\right| - 3 \) 4. \( i\left( x\right) = 4 - 2\left| {{3x} + 1}\right| \)\n\nFind the zeros of each function and the \( x \) - and \( y \) -intercepts o... | ## Solution.\n\n1. To find the zeros of \( f \), we set \( f\left( x\right) = 0 \) . We get \( \left| x\right| = 0 \), which, by Theorem 2.1 gives us \( x = 0 \) . Since the zeros of \( f \) are the \( x \) -coordinates of the \( x \) -intercepts of the graph of \( y = f\left( x\right) \), we get \( \left( {0,0}\right)... | Yes |
Graph the following functions starting with the graph of \( f\left( x\right) = \left| x\right| \) and using transformations. | Solution. We begin by graphing \( f\left( x\right) = \left| x\right| \) and labeling three points, \( \left( {-1,1}\right) ,\left( {0,0}\right) \) and \( \left( {1,1}\right) \) .\n\n\n\n\( f\left( x\right) = \left| x... | Yes |
1. \( f\left( x\right) = \frac{\left| x\right| }{x} \) | We first note that, due to the fraction in the formula of \( f\left( x\right), x \neq 0 \) . Thus the domain is \( \left( {-\infty ,0}\right) \cup \left( {0,\infty }\right) \) . To find the zeros of \( f \), we set \( f\left( x\right) = \frac{\left| x\right| }{x} = 0 \) . This last equation implies \( \left| x\right| =... | Yes |
Graph the following functions starting with the graph of \( f\left( x\right) = {x}^{2} \) and using transformations. Find the vertex, state the range and find the \( x \) - and \( y \) -intercepts, if any exist.\n\n1. \( g\left( x\right) = {\left( x + 2\right) }^{2} - 3 \) | Since \( g\left( x\right) = {\left( x + 2\right) }^{2} - 3 = f\left( {x + 2}\right) - 3 \), Theorem 1.7 instructs us to first subtract 2 from each of the \( x \) -values of the points on \( y = f\left( x\right) \) . This shifts the graph of \( y = f\left( x\right) \) to the left 2 units and moves \( \left( {-2,4}\right... | Yes |
Theorem 2.2. Vertex Formula for Quadratics in Standard Form: For the quadratic function \( f\left( x\right) = a{\left( x - h\right) }^{2} + k \), where \( a, h \) and \( k \) are real numbers with \( a \neq 0 \), the vertex of the graph of \( y = f\left( x\right) \) is \( \left( {h, k}\right) \) . | To see why the formula in Theorem 2.2 produces the vertex, consider the graph of the equation \( y = a{\left( x - h\right) }^{2} + k \) . When we substitute \( x = h \), we get \( y = k \), so \( \left( {h, k}\right) \) is on the graph. If \( x \neq h \), then \( x - h \neq 0 \) so \( {\left( x - h\right) }^{2} \) is a... | Yes |
Convert the functions below from general form to standard form. Find the vertex, axis of symmetry and any \( x \) - or \( y \) -intercepts. Graph each function and determine its range. | 1. To convert from general form to standard form, we complete the square. \( {}^{7} \) First, we verify that the coefficient of \( {x}^{2} \) is 1 . Next, we find the coefficient of \( x \), in this case -4, and take half of it to get \( \frac{1}{2}\left( {-4}\right) = - 2 \) . This tells us that our target perfect squ... | Yes |
1. Determine the weekly profit function \( P\left( x\right) \) . | 1. To find the profit function \( P\left( x\right) \), we subtract\n\n\[ P\left( x\right) = R\left( x\right) - C\left( x\right) = \left( {-{1.5}{x}^{2} + {250x}}\right) - \left( {{80x} + {150}}\right) = - {1.5}{x}^{2} + {170x} - {150}. \]\n\nSince the revenue function is valid when \( 0 \leq x \leq {166}, P \) is also ... | Yes |
Much to Donnie's surprise and delight, he inherits a large parcel of land in Ashtabula County from one of his (e)strange(d) relatives. The time is finally right for him to pursue his dream of farming alpaca. He wishes to build a rectangular pasture, and estimates that he has enough money for 200 linear feet of fencing ... | Solution. It is always helpful to sketch the problem situation, so we do so below.\n\n\n\nWe are tasked to find the dimensions of the pasture which would give a maximum area. We let \( w \) denote the width of the pa... | Yes |
Example 2.3.5. Graph \( f\left( x\right) = \left| {{x}^{2} - x - 6}\right| \) . | Solution. Using the definition of absolute value, Definition 2.4, we have\n\n\[ f\left( x\right) = \left\{ \begin{array}{rrr} - \left( {{x}^{2} - x - 6}\right) , & \text{ if } & {x}^{2} - x - 6 < 0 \\ {x}^{2} - x - 6, & \text{ if } & {x}^{2} - x - 6 \geq 0 \end{array}\right. \]\n\nThe trouble is that we have yet to dev... | No |
1. Solve \( f\left( x\right) = g\left( x\right) \) . | 1. To solve \( f\left( x\right) = g\left( x\right) \), we replace \( f\left( x\right) \) with \( {2x} - 1 \) and \( g\left( x\right) \) with 5 to get \( {2x} - 1 = 5 \) . Solving for \( x \), we get \( x = 3 \) . | Yes |
1. Solve \( f\left( x\right) = g\left( x\right) \) . | 1. To solve \( f\left( x\right) = g\left( x\right) \), we look for where the graphs of \( f \) and \( g \) intersect. These appear to be at the points \( \left( {-1,2}\right) \) and \( \left( {1,2}\right) \), so our solutions to \( f\left( x\right) = g\left( x\right) \) are \( x = - 1 \) and \( x = 1 \) . | Yes |
Theorem 2.4. Inequalities Involving the Absolute Value: Let \( c \) be a real number.\n\n- For \( c > 0,\left| x\right| < c \) is equivalent to \( - c < x < c \) . | As with Theorem 2.1 in Section 2.2, we could argue Theorem 2.4 using cases. However, in light of what we have developed in this section, we can understand these statements graphically. For instance, if \( c > 0 \), the graph of \( y = c \) is a horizontal line which lies above the \( x \) -axis through \( \left( {0, c}... | Yes |
1. \( \left| {x - 1}\right| \geq 3 \) | From Theorem 2.4, \( \left| {x - 1}\right| \geq 3 \) is equivalent to \( x - 1 \leq - 3 \) or \( x - 1 \geq 3 \) . Solving, we get \( x \leq - 2 \) or \( x \geq 4 \), which, in interval notation is \( \left( {-\infty , - 2\rbrack \cup \lbrack 4,\infty }\right) \) . Graphically, we have\n\n\n\nWe can tell both from the correlation coefficient as well as the graph that the regression line is a good fit to the data. | No |
Example 2.5.2. Using the quadratic model for the temperature data above, predict the warmest temperature of the day. When will this occur? | Solution. The maximum temperature will occur at the vertex of the parabola. Recalling the Vertex Formula, Equation 2.4, \( x = - \frac{b}{2a} \approx - \frac{9.464}{2\left( {-{0.321}}\right) } \approx {14.741} \) . This corresponds to roughly 2: 45 PM. To find the temperature, we substitute \( x = {14.741} \) into \( y... | Yes |
Example 3.1.1. Determine if the following functions are polynomials. Explain your reasoning. | 1. We note directly that the domain of \( g\left( x\right) = \frac{{x}^{3} + 4}{x} \) is \( x \neq 0 \) . By definition, a polynomial has all real numbers as its domain. Hence, \( g \) can’t be a polynomial.\n\n2. Even though \( p\left( x\right) = \frac{{x}^{3} + {4x}}{x} \) simplifies to \( p\left( x\right) = {x}^{2} ... | Yes |
Find the degree, leading term, leading coefficient and constant term of the following polynomial functions. | ## Solution.\n\n1. There are no surprises with \( f\left( x\right) = 4{x}^{5} - 3{x}^{2} + {2x} - 5 \) . It is written in the form of Definition 3.2, and we see that the degree is 5, the leading term is \( 4{x}^{5} \), the leading coefficient is 4 and the constant term is -5 .\n\n2. The form given in Definition 3.2 has... | Yes |
A box with no top is to be fashioned from a 10 inch \( \times {12} \) inch piece of cardboard by cutting out congruent squares from each corner of the cardboard and then folding the resulting tabs. Let \( x \) denote the length of the side of the square which is removed from each corner. \n\nSince the dividend was a third degree polynomial, the quotient is a quadrati... | Yes |
1. Find \( p\left( {-2}\right) \) using The Remainder Theorem. Check your answer by substitution. | 1. The Remainder Theorem states \( p\left( {-2}\right) \) is the remainder when \( p\left( x\right) \) is divided by \( x - \left( {-2}\right) \) . We set up our synthetic division tableau below. We are careful to record the coefficient of \( {x}^{2} \) as 0 , and proceed as above.\n\n\n\nFrom the first divi... | Yes |
Theorem 3.7. Suppose \( f \) is a polynomial of degree \( n \geq 1 \) . Then \( f \) has at most \( n \) real zeros, counting multiplicities. | Theorem 3.7 is a consequence of the Factor Theorem and polynomial multiplication. Every zero \( c \) of \( f \) gives us a factor of the form \( \left( {x - c}\right) \) for \( f\left( x\right) \) . Since \( f \) has degree \( n \), there can be at most \( n \) of these factors. | Yes |
Theorem 3.8. Cauchy’s Bound: Suppose \( f\left( x\right) = {a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \ldots + {a}_{1}x + {a}_{0} \) is a polynomial of degree \( n \) with \( n \geq 1 \) . Let \( M \) be the largest of the numbers: \( \frac{\left| {a}_{0}\right| }{\left| {a}_{n}\right| },\frac{\left| {a}_{1}\right| }{\l... | The proof of this fact is not easily explained within the confines of this text. This paper contains the result and gives references to its proof. | No |
Example 3.3.1. Let \( f\left( x\right) = 2{x}^{4} + 4{x}^{3} - {x}^{2} - {6x} - 3 \) . Determine an interval which contains all of the real zeros of \( f \) . | Solution. To find the \( M \) stated in Cauchy’s Bound, we take the absolute value of the leading coefficient, in this case \( \left| 2\right| = 2 \) and divide it into the largest (in absolute value) of the remaining coefficients, in this case \( \left| {-6}\right| = 6 \) . This yields \( M = 3 \) so it is guaranteed ... | Yes |
Theorem 3.9. Rational Zeros Theorem: Suppose \( f\left( x\right) = {a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \ldots + {a}_{1}x + {a}_{0} \) is a polynomial of degree \( n \) with \( n \geq 1 \), and \( {a}_{0},{a}_{1},\ldots {a}_{n} \) are integers. If \( r \) is a rational zero of \( f \), then \( r \) is of the form ... | The Rational Zeros Theorem gives us a list of numbers to try in our synthetic division and that is a lot nicer than simply guessing. If none of the numbers in the list are zeros, then either the polynomial has no real zeros at all, or all of the real zeros are irrational numbers. To see why the Rational Zeros Theorem w... | Yes |
Let \( f\left( x\right) = 2{x}^{4} + 4{x}^{3} - {x}^{2} - {6x} - 3 \) . Use the Rational Zeros Theorem to list all of the possible rational zeros of \( f \) . | Solution. To generate a complete list of rational zeros, we need to take each of the factors of constant term, \( {a}_{0} = - 3 \), and divide them by each of the factors of the leading coefficient \( {a}_{4} = 2 \) . The factors of \( - 3 \) are \( \pm 1 \) and \( \pm 3 \) . Since the Rational Zeros Theorem tacks on a... | Yes |
Let \( f\left( x\right) = 2{x}^{4} + 4{x}^{3} - {x}^{2} - {6x} - 3 \) .\n\n1. Graph \( y = f\left( x\right) \) on the calculator using the interval obtained in Example 3.3.1 as a guide.\n\n2. Use the graph to shorten the list of possible rational zeros obtained in Example 3.3.2.\n\n3. Use synthetic division to find the... | ## Solution.\n\n1. In Example 3.3.1, we determined all of the real zeros of \( f \) lie in the interval \( \left\lbrack {-4,4}\right\rbrack \) . We set our window accordingly and get\n\n\n\n2. In Example 3.3.2, we le... | Yes |
Example 3.3.4. Let \( f\left( x\right) = {x}^{4} + {x}^{2} - {12} \). | ## Solution.\n\n1. Applying Cauchy’s Bound, we find \( M = {12} \), so all of the real zeros lie in the interval \( \left\lbrack {-{13},{13}}\right\rbrack \) .\n\n2. Applying the Rational Zeros Theorem with constant term \( {a}_{0} = - {12} \) and leading coefficient \( {a}_{4} = 1 \), we get the list \( \{ \pm 1, \pm ... | Yes |
Let \( f\left( x\right) = 2{x}^{4} + 4{x}^{3} - {x}^{2} - {6x} - 3 \) . Use Descartes’ Rule of Signs to determine the possible number and location of the real zeros of \( f \) . | Solution. As noted above, the variations of sign of \( f\left( x\right) \) is 1 . This means, counting multiplicities, \( f \) has exactly 1 positive real zero. Since \( f\left( {-x}\right) = 2{\left( -x\right) }^{4} + 4{\left( -x\right) }^{3} - {\left( -x\right) }^{2} - 6\left( {-x}\right) - 3 = \) \( 2{x}^{4} - 4{x}^... | Yes |
Theorem 3.11. Upper and Lower Bounds: Suppose \( f \) is a polynomial of degree \( n \geq 1 \) . - If \( c > 0 \) is synthetically divided into \( f \) and all of the numbers in the final line of the division tableau have the same signs, then \( c \) is an upper bound for the real zeros of \( f \) . That is, there are ... | The Upper and Lower Bounds Theorem works because of Theorem 3.4. For the upper bound part of the theorem, suppose \( c > 0 \) is divided into \( f \) and the resulting line in the division tableau contains, for example, all nonnegative numbers. This means \( f\left( x\right) = \left( {x - c}\right) q\left( x\right) + r... | Yes |
1. Find all of the real zeros of \( f \) and their multiplicities. | We know from Cauchy’s Bound that all of the real zeros lie in the interval \( \left\lbrack {-4,4}\right\rbrack \) and that our possible rational zeros are \( \pm \frac{1}{2}, \pm 1, \pm \frac{3}{2} \) and \( \pm 3 \) . Descartes’ Rule of Signs guarantees us at least one negative real zero and exactly one positive real ... | Yes |
Example 3.3.8. Suppose the profit \( P \), in thousands of dollars, from producing and selling \( x \) hundred LCD TVs is given by \( P\left( x\right) = - 5{x}^{3} + {35}{x}^{2} - {45x} - {25},0 \leq x \leq {10.07} \) . How many TVs should be produced to make a profit? | Solution. To ’make a profit’ means to solve \( P\left( x\right) = - 5{x}^{3} + {35}{x}^{2} - {45x} - {25} > 0 \), which we do analytically using a sign diagram. To simplify things, we first factor out the -5 common to all the coefficients to get \( - 5\left( {{x}^{3} - 7{x}^{2} + {9x} - 5}\right) > 0 \), so we can just... | Yes |
Perform the indicated operations. Write your answer in the form \( {}^{5}a + {bi} \) . | 1. As mentioned earlier, we treat expressions involving \( i \) as we would any other radical. We combine like terms to get \( \left( {1 - {2i}}\right) - \left( {3 + {4i}}\right) = 1 - {2i} - 3 - {4i} = - 2 - {6i} \) . | No |
Theorem 3.12. Properties of the Complex Conjugate: Let \( z \) and \( w \) be complex numbers.\n\n- \( \overline{\bar{z}} = z \)\n\n- \( \bar{z} + \bar{w} = \overline{z + w} \)\n\n- \( \bar{z}\bar{w} = \overline{zw} \)\n\n- \( {\left( \bar{z}\right) }^{n} = \overline{{z}^{n}} \), for any natural number \( n \)\n\n- \( ... | Essentially, Theorem 3.12 says that complex conjugation works well with addition, multiplication and powers. The proof of these properties can best be achieved by writing out \( z = a + {bi} \) and \( w = c + {di} \) for real numbers \( a, b, c \) and \( d \) . Next, we compute the left and right hand sides of each equ... | No |
Find all of the complex zeros of \( f \) and state their multiplicities. | Since \( f \) is a fifth degree polynomial, we know that we need to perform at least three successful divisions to get the quotient down to a quadratic function. At that point, we can find the remaining zeros using the Quadratic Formula, if necessary. Using the techniques developed in Section 3.3, we get\n\n | No |
Find a polynomial \( p \) of lowest degree that has integer coefficients and satisfies all of the following criteria:\n\n- the graph of \( y = p\\left( x\\right) \) touches (but doesn’t cross) the \( x \) -axis at \( \\left( {\\frac{1}{3},0}\\right) \)\n\n- \( x = {3i} \) is a zero of \( p \).\n\n- as \( x \\rightarrow... | Solution. To solve this problem, we will need a good understanding of the relationship between the \( x \) -intercepts of the graph of a function and the zeros of a function, the Factor Theorem, the role of multiplicity, complex conjugates, the Complex Factorization Theorem, and end behavior of polynomial functions. (I... | Yes |
Find the domain of the following rational functions. Write them in the form \( \frac{p\left( x\right) }{q\left( x\right) } \) for polynomial functions \( p \) and \( q \) and simplify. | 1. To find the domain of \( f \), we proceed as we did in Section 1.4: we find the zeros of the denominator and exclude them from the domain. Setting \( x + 1 = 0 \) results in \( x = - 1 \) . Hence, our domain is \( \left( {-\infty , - 1}\right) \cup \left( {-1,\infty }\right) \) . The expression \( f\left( x\right) \... | Yes |
Find the vertical asymptotes of, and/or holes in, the graphs of the following rational functions. Verify your answers using a graphing calculator, and describe the behavior of the graph near them using proper notation. | 1. To use Theorem 4.1, we first find all of the real numbers which aren’t in the domain of \( f \) . To do so, we solve \( {x}^{2} - 3 = 0 \) and get \( x = \pm \sqrt{3} \) . Since the expression \( f\left( x\right) \) is in lowest terms, there is no cancellation possible, and we conclude that the lines \( x = - \sqrt{... | Yes |
1. Find and interpret \( P\left( 0\right) \) . | Substituting \( t = 0 \) gives \( P\left( 0\right) = \frac{100}{{\left( 5 - 0\right) }^{2}} = 4 \), which means 4000 bacteria are initially introduced into the environment. | Yes |
List the horizontal asymptotes, if any, of the graphs of the following functions. Verify your answers using a graphing calculator, and describe the behavior of the graph near them using proper notation. | 1. The numerator of \( f\left( x\right) \) is \( {5x} \), which has degree 1 . The denominator of \( f\left( x\right) \) is \( {x}^{2} + 1 \), which has degree 2. Applying Theorem 4.2, \( y = 0 \) is the horizontal asymptote. Sure enough, we see from the graph that as \( x \rightarrow - \infty, f\left( x\right) \righta... | Yes |
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