Answer:
[tex]c = \frac{-2}{3}[/tex]
Step-by-step explanation:
We are given the following:
[tex]f(x)=24x^2+16x+7[/tex]
Interval: [-1,1]
Mean Value theorem:
If the given function is continuous n [a,b] and differentiable in (a,b), then there exist a c in ((a,b) such that
[tex]f'(c) = \displaystyle\frac{f(b)-f(a)}{b-a}[/tex]
Now, we evaluate f(-1), f(1) and f'(x)
[tex]f'(x) = 48x+16\\f(1)= 24(1)^2+16(1)+7 = 47\\f(-1) = 24(-1)^2+16(-1)+7 = 15[/tex]
By mean value theorem:
[tex]f'(c) = \displaystyle\frac{f(b)-f(a)}{b-a}\\\\48c + 16 = \frac{15-47}{1-(-1)}\\\\48c + 16 = -16\\48c = -32\\\\c = \frac{-32}{48} = \frac{-2}{3}\\\\[/tex]
