Verify Rolle's theorem for the function y=x^{ | vedclass

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Question

(A) The function $f(x) = x^{2} + 2$ is a polynomial function,which is continuous on the closed interval $[-2, 2]$ and differentiable on the open inter

Vedclass · Accepted Answer

The theorem is verified at $c=0$.

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(A) The function $f(x) = x^{2} + 2$ is a polynomial function,which is continuous on the closed interval $[-2, 2]$ and differentiable on the open interval $(-2, 2)$.Next,we calculate the values at the endpoints:$f(-2) = (-2)^{2} + 2 = 4 + 2 = 6$$f(2) = (2)^{2} + 2 = 4 + 2 = 6$Since $f(-2) = f(2) = 6$,all conditions of Rolle's theorem are satisfied.According to Rolle's theorem,there exists at least one point $c \in (-2, 2)$ such that $f'(c) = 0$.We find the derivative: $f'(x) = 2x$.Setting $f'(c) = 0$,we get $2c = 0$,which implies $c = 0$.Since $0 \in (-2, 2)$,Rolle's theorem is verified at $c = 0$.

Verify Rolle's theorem for the function $y=x^{2}+2$ on the interval $[-2, 2]$.

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