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Verify Rolle’s Theorem for the function $y = {x^2} + 2$ where $x \in \left[ { - 2,2} \right].$

Last updated date: 26th Mar 2023
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Hint: Rolle’s Theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a)=f(b), then$f'(x) = 0$ for some x, $a \leqslant x \leqslant b$.

Given function is $y = {x^2} + 2$ and x belongs to [-2, 2]
Here y is a function of x $ \Rightarrow y = f(x)$
The given function 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 Rolle’s 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 $0 \in \left( { - 2,2} \right)$ such that $f'(0) = 0$
Therefore, Rolle’s Theorem is verified.

Note:While verifying Rolle’s Theorem for a function, we need to make sure that it is satisfying all the three rules of Rolle’s Theorem. If any of those conditions failed then, we can say that Rolle’s Theorem is not applicable for that function.
Function is continuous means its graph is unbroken without any holes (discontinuity). Function must be defined at every point of the given interval.