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Question

Verify Rolle's theorem for the function $$y=x^2+2$$. $$x\epsilon |-2, 2|$$.


Solution

Given that $$y=x^{2}+2$$ and $$x$$ belongs to $$[-2,2]$$
Here $$f(x)$$ is continuous on a closed interval $$[-2,2]$$ and differentiable on an open interval $$(-2,2)$$
We have $$f(2)=f(-2)=6$$
According to rolles theorem , if $$f(-2)=f(2)$$ then there exists at least one point $$c$$ in $$(-2,2)$$ such that $$f'{(c)}=0$$
Now, to check whether such $$c$$ exists or not
We have $$f'{(x)}=2x$$
$$f'{(x)}=2x=0$$ for $$x=0$$, and $$-2<0<2$$
Hence, there exist $$0\in (-2,2)$$ such that $$f'(0)=0$$
Therefore, Rolle's theorem is verified.

Mathematics

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