5 7Consider the function y =5x2 – 5x + 9 on the interval2a) Find the slope of the secant line on this interval.m =b) Find the value(s) for c that satisfy the Rolle's Theorem on the given interval.C =

Given: y=5x2-5x+9 defined on the interval -52,72. To Find: a. Find the slope of the secant line on…

Answered: 5 7 Consider the function y = V5x? 5x +…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter3: Functions And Graphs

Section3.3: Lines

Problem 36E

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5 7
Consider the function y =
5x2 – 5x + 9 on the interval
2
a) Find the slope of the secant line on this interval.
m =
b) Find the value(s) for c that satisfy the Rolle's Theorem on the given interval.
C =

Expert Solution

Step 1

Given:

$y = \sqrt{5 x^{2} - 5 x + 9} defined on the interval [- \frac{5}{2}, \frac{7}{2}] .$

To Find:

$(a) . Find the slope of the secant line on the given interval. (b) . Find the value (s) for c that satisfy the Rolle's Theorem on the given interval.$

Step 2

Explanation:

$(a) . Slope of the secant line is given by: m_{s e c} = \frac{f (x + ∆ x) - f (x)}{∆ x} ∆ x is the distance to the end of the interval. here, we have y = f (x) = \sqrt{5 x^{2} - 5 x + 9} defined on the interval [- \frac{5}{2}, \frac{7}{2}] x : is the start of the interval \Rightarrow x = - \frac{5}{2} and ∆ x = \frac{7}{2} - (- \frac{5}{2}) = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6 \Rightarrow ∆ x = 6 \Rightarrow m_{s e c} = \frac{f (- \frac{5}{2} + 6) - f (- \frac{5}{2})}{6} = \frac{f (\frac{7}{2}) - f (- \frac{5}{2})}{6} \Rightarrow m_{s e c} = \frac{f (\frac{7}{2}) - f (- \frac{5}{2})}{6} = \frac{\sqrt{5 {(\frac{7}{2})}^{2} - 5 (\frac{7}{2}) + 9} - \sqrt{5 {(- \frac{5}{2})}^{2} - 5 (\frac{5}{2}) + 9}}{6} \Rightarrow m_{s e c} = \frac{\sqrt{5 (\frac{49}{4}) - 5 (\frac{7}{2}) + 9} - \sqrt{5 (\frac{25}{4}) - 5 (\frac{5}{2}) + 9}}{6} \Rightarrow m_{s e c} = \frac{\sqrt{\frac{245}{4} - \frac{35}{2} + 9} - \sqrt{\frac{125}{4} - \frac{25}{2} + 9}}{6} \Rightarrow m_{s e c} = \frac{\sqrt{\frac{245 - 70 + 36}{4}} - \sqrt{\frac{125 - 50 + 36}{4}}}{6} \Rightarrow m_{s e c} = \frac{\sqrt{\frac{211}{4}} - \sqrt{\frac{111}{4}}}{6} = \frac{\sqrt{211} - \sqrt{111}}{12} = 0.3325 \Rightarrow m_{s e c} = 0.3325$

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5 7 Consider the function y = V5x? 5x + 9 on the interval - - 2' 2 a) Find the slope of the secant line on this interval. m = b) Find the value(s) for c that satisfy the Rolle's Theorem on the given interval. C =