If f(x) is differentiable for the closed interval [−4, 0] such that f(−4) = 5 and f(0) = 9, then there exists a value c, −4 <

c < 0 such that
f(c) = 0
f '(c) = 0
f (c) = 1
f '(c) = 1

Mathematics

1 answer:

Pavel [41]2 years ago

4 0

$\bf \textit{mean value theorem}\\\\
f'(c)=\cfrac{f(b)-f(a)}{b-a}\qquad 
\begin{cases}
a=-4\\
b=0
\end{cases}\implies f'(c)=\cfrac{f(0)-f(-4)}{0-(-4)}
\\\\\\
f'(c)=\cfrac{9-5}{0+4}\implies f'(c)=\cfrac{4}{4}\implies f'(c)=1$

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