Question

What is the value of c in the interval (5,8) guaranteed by Rolle's Theorem for the function g(x)=−7x3+91x2−280x−9? Note that g(5)=g(8)=−9. (Do not include "c=" in your answer.)

Answer 1

Answer:

$\displaystyle c = \frac{20}{3}$

Step-by-step explanation:

According to Rolle's Theorem, if f(a) = f(b) in an interval [a, b], then there must exist at least one c within (a, b) such that f'(c) = 0.

We are given that g(5) = g(8) = -9. Then according to Rolle's Theorem, there must be a c in (5, 8) such that g'(c) = 0.

So, differentiate the function. We can take the derivative of both sides with respect to x:

$\displaystyle g'(x) = \frac{d}{dx}\left[ -7x^3 +91x^2 -280x - 9\right]$

Differentiate:

$g'(x) = -21x^2+182x-280$

Let g'(x) = 0:

$0 = -21x^2+182x-280$

Solve for x. First, divide everything by negative seven:

$0=3x^2-26x+40$

Factor:

$0=(x-2)(3x-20)$

Zero Product Property:

$x-2=0 \text{ or } 3x-20=0$

Solve for each case. Hence:

$\displaystyle x=2 \text{ or } x = \frac{20}{3}$

Since the first solution is not within our interval, we can ignore it.

Therefore:

$\displaystyle c = \frac{20}{3}$