Answer:
[tex]\displaystyle c = \frac{20}{3}[/tex]
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:
[tex]\displaystyle g'(x) = \frac{d}{dx}\left[ -7x^3 +91x^2 -280x - 9\right][/tex]
Differentiate:
[tex]g'(x) = -21x^2+182x-280[/tex]
Let g'(x) = 0:
[tex]0 = -21x^2+182x-280[/tex]
Solve for x. First, divide everything by negative seven:
[tex]0=3x^2-26x+40[/tex]
Factor:
Zero Product Property:
[tex]x-2=0 \text{ or } 3x-20=0[/tex]
Solve for each case. Hence:
[tex]\displaystyle x=2 \text{ or } x = \frac{20}{3}[/tex]
Since the first solution is not within our interval, we can ignore it.
Therefore:
[tex]\displaystyle c = \frac{20}{3}[/tex]
