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gizmo_the_mogwai [7]
3 years ago
9

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]3 years ago
4 0
\bf \textit{mean value theorem}\\\\&#10;f'(c)=\cfrac{f(b)-f(a)}{b-a}\qquad &#10;\begin{cases}&#10;a=-4\\&#10;b=0&#10;\end{cases}\implies f'(c)=\cfrac{f(0)-f(-4)}{0-(-4)}&#10;\\\\\\&#10;f'(c)=\cfrac{9-5}{0+4}\implies f'(c)=\cfrac{4}{4}\implies f'(c)=1
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Step-by-step explanation:

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2 years ago
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