Medium
Verify Rolle's Theorem for the function $f(x)=x^{2}+2x-8, x \in[-4,2]$
(N/A) The given function $f(x)=x^{2}+2x-8$ is a polynomial function,which is continuous on the closed interval $[-4,2]$ and differentiable on the open interval $(-4,2).$
First,we evaluate the function at the endpoints:
$f(-4) = (-4)^{2} + 2(-4) - 8 = 16 - 8 - 8 = 0$
$f(2) = (2)^{2} + 2(2) - 8 = 4 + 4 - 8 = 0$
Since $f(-4) = f(2) = 0$,the conditions for Rolle's Theorem are satisfied.
According to Rolle's Theorem,there exists at least one point $c \in (-4,2)$ such that $f'(c) = 0.$
We find the derivative of $f(x)$:
$f'(x) = \frac{d}{dx}(x^{2} + 2x - 8) = 2x + 2$
Setting $f'(c) = 0$:
$2c + 2 = 0$
$2c = -2$
$c = -1$
Since $c = -1$ lies in the interval $(-4,2)$,Rolle's Theorem is verified.
First,we evaluate the function at the endpoints:
$f(-4) = (-4)^{2} + 2(-4) - 8 = 16 - 8 - 8 = 0$
$f(2) = (2)^{2} + 2(2) - 8 = 4 + 4 - 8 = 0$
Since $f(-4) = f(2) = 0$,the conditions for Rolle's Theorem are satisfied.
According to Rolle's Theorem,there exists at least one point $c \in (-4,2)$ such that $f'(c) = 0.$
We find the derivative of $f(x)$:
$f'(x) = \frac{d}{dx}(x^{2} + 2x - 8) = 2x + 2$
Setting $f'(c) = 0$:
$2c + 2 = 0$
$2c = -2$
$c = -1$
Since $c = -1$ lies in the interval $(-4,2)$,Rolle's Theorem is verified.
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