Verify Rolle's Theorem for the function $f(x)=x^{2}+2 x-8, x \in[-4,2]$
Verify Rolle's Theorem for the function $f(x)=x^{2}+2 x-8, x \in[-4,2]$
The given function, $f(x)=x^{2}+2 x-8,$ being polynomial function, is continuous in $[-4,2]$ and is differentiable in $(-4,2).$
$f(-4)=(-4)^{2}+2 x(-4)-8=16-8-8=0$
$f(2)=(2)^{2}+2 \times 2-8=4+4-8=0$
$\therefore f(-4)=f(2)=0$
$\Rightarrow$ The value of $f(x)$ at $-4$ and $2$ coincides.
Rolle's Theorem states that there is a point $c \in(-4,2)$ such that $f^{\prime}(c)=0$
$f(x)=x^{2}+2 x-8$
$\Rightarrow f^{\prime}(x)=2 x+2$
$\therefore f^{\prime}(c)=0$
$\Rightarrow 2 c+2=-1$
$\Rightarrow c=-1$
$c=-1 \in(-4,2)$
Hence, Rolle's Theorem is verified for the given function.
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