Explanation Given Information: The functions are ( i ) f ( x ) = 2 x + 3 at ( 1 , 5 ) . ( ii ) f ( x ) = x 2 + 4 at ( − 1 , 5 ) . ( iii ) f ( x ) = 5 x 2 at ( − 2 , 20 ) . ( iv ) f ( x ) = 2 x 2 − x at ( 2 , 6 ) . ( v ) f ( x ) = 2 x 2 − x + 5 at ( 0 , 5 ) . Graph: ( i ) Consider the function f ( x ) = 2 x + 3 and the point ( 1 , 5 ) Here, ( a , f ( a ) ) = ( 1 , 5 ) Now, compute the slope of the tangent line for the function f ( x ) = 2 x + 3 as follows: Put a = 1 , m tan = lim h → 0 f ( 1 + h ) − f ( 1 ) h To compute f ( 1 + h ) , substitute x = 1 + h in the function f ( x ) = 2 x + 3 . m tan = lim h → 0 [ 2 ( 1 + h ) + 3 ] − [ 2 ( 1 ) + 3 ] h Now, simplify 2 ( 1 + h ) by using the distributive property A ( B + C ) = A B + A C . m tan = lim h → 0 [ 2 + 2 h + 3 ] − 5 h Combine the like terms in the numerator. m tan = lim h → 0 2 h h Cancel out h and apply the limits. m tan = lim h → 0 2 = 2 Thus, the slope of the tangent to the graph of f ( x ) = 2 x + 3 at ( 1 , 5 ) is 2 . Now compute the slope intercept equation of the tangent line as follows: Here, the tangent line passes through the point ( 1 , 5 ) which implies x 1 = 1 and y 1 = 5 Substitute m = 2 , x 1 = 1 , y 1 = 5 in the equation y − y 1 = m ( x − x 1 ) . y − 5 = 2 ( x − 1 ) y − 5 = 2 x − 2 y = 2 x − 2 + 5 y = 2 x + 3 Now, graph the function and the tangent using the TI- 83 graphing calculator as follows: Step1. Press “ Y = ” key and insert the function and the tangent as, Y 1 = 2 X + 3 Y 2 = 2 X + 3 Step2. Press the “WINDOW” key and set the values as, X min = − 10 , Xmax = 10 and Xscl = 1 And Ymin = − 10 , Ymax = 10 and Yscl = 1 . Step3. Then press the “GRAPH” key. A window showing the graph of the function and the tangent will appear as, The graph of the function f ( x ) = 2 x + 3 is a line and there is no tangent drawn in the graph because a tangent line to a straight line is the line itself. ( ii ) Consider the function f ( x ) = x 2 + 4 and the point ( − 1 , 5 ) Here, ( a , f ( a ) ) = ( − 1 , 5 ) Now, compute the slope of the tangent line for the function f ( x ) = x 2 + 4 as follows: Put a = − 1 , m tan = lim h → 0 f ( − 1 + h ) − f ( − 1 ) h To compute f ( − 1 + h ) , substitute x = − 1 + h in the function f ( x ) = x 2 + 4 . m tan = lim h → 0 [ ( − 1 + h ) 2 + 4 ] − [ ( − 1 ) 2 + 4 ] h Now, simplify ( − 1 + h ) 2 by using the property ( A + B ) 2 = A 2 + 2 A B + B 2 . m tan = lim h → 0 [ 1 − 2 h + h 2 + 4 ] − 5 h Combine the like terms in the numerator. m tan = lim h → 0 h 2 − 2 h h Now, factor the numerator and divide the numerator as well as the denominator by h , m tan = lim h → 0 h ( h − 2 ) h = lim h → 0 ( h − 2 ) Apply the limits and simplify. m tan = 0 − 2 = − 2 Thus, the slope of the tangent to the graph of f ( x ) = x 2 + 4 at ( − 1 , 5 ) is − 2 . Now compute the slope intercept equation of the tangent line as follows: Here, the tangent line passes through the point ( − 1 , 5 ) which implies x 1 = − 1 and y 1 = 5 Substitute m = − 2 , x 1 = − 1 , y 1 = 5 in the equation y − y 1 = m ( x − x 1 ) . y − 5 = − 2 [ x − ( − 1 ) ] y − 5 = − 2 ( x + 1 ) y = − 2 x − 2 + 5 y = − 2 x + 3 Now, graph the function and the tangent using the TI- 83 graphing calculator as follows: Step1. Press “ Y = ” key and insert the function and the tangent as, Y 1 = X ^ 2 + 4 Y 2 = − 2X + 3 Step2. Press the “WINDOW” key and set the values as, X min = − 10 , Xmax = 10 and Xscl = 1 And Ymin = − 10 , Ymax = 10 and Yscl = 1 . Step3. Then press the “GRAPH” key. A window showing the graph of the function and tangent, will appear as, The line y = − 2 x + 3 is the tangent line to the graph of f ( x ) = x 2 + 4 at ( − 1 , 5 ) . ( iii ) Consider the function f ( x ) = 5 x 2 and the point ( − 2 , 20 ) Here, ( a , f ( a ) ) = ( − 2 , 20 ) Now, compute the slope of the tangent line for the function f ( x ) = 5 x 2 as follows: Put a = − 2 , m tan = lim h → 0 f ( − 2 + h ) − f ( − 2 ) h To compute f ( − 2 + h ) , substitute x = − 2 + h in the function f ( x ) = 5 x 2 . m tan = lim h → 0 [ 5 ( − 2 + h ) 2 ] − 5 ( − 2 ) 2 h Now, simplify ( − 2 + h ) 2 by using the property ( A − B ) 2 = A 2 − 2 A B + B 2 . m tan = lim h → 0 5 ( 4 − 4 h + h 2 ) − 20 h = lim h → 0 20 − 20 h + 5 h 2 − 20 h Combine the like terms in the numerator. m tan = lim h → 0 5 h 2 − 20 h h Now, factor the numerator and divide the numerator as well as the denominator by h , m tan = lim h → 0 5 h ( h − 4 ) h = lim h → 0 5 ( h − 4 ) Apply the limits and simplify

6th Edition

ISBN: 9780134469140

Author: Robert F. Blitzer

Publisher: PEARSON

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Chapter 11.4, Problem 60PE

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To graph: Any five functions and their tangent lines without using the DRAW TANGENT feature of the graphing utility and check if the line appears to be tangent to the graph of f at the given point on f.

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Chapter 11.4, Problem 59PE

Chapter 11.4, Problem 61PE

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Chapter 11 Solutions

Precalculus (6th Edition)

Ch. 11.1 - Check Point 1 Find: . Ch. 11.1 - Prob. 2CP Ch. 11.1 - Prob. 3CP Ch. 11.1 - Prob. 4CP Ch. 11.1 - Prob. 5CP Ch. 11.1 - Prob. 1CVC Ch. 11.1 - Prob. 2CVC Ch. 11.1 - Prob. 3CVC Ch. 11.1 - Fill in each blank so that the resulting statement...Ch. 11.1 - Fill in each blank so that the resulting statement...

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To graph: Any five functions and their tangent lines without using the DRAW TANGENT feature of the graphing utility and check if the line appears to be tangent to the graph of f at the given point on f .

Solution Summary: The author explains how to graph any five functions and their tangent lines without using the DRAW TANGENT feature of the graphing utility.

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To graph: Any five functions and their tangent lines without using the DRAW TANGENT feature of the graphing utility and check if the line appears to be tangent to the graph of f at the given point on f.

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Chapter 11 Solutions