← Back to learning objective

Verifying MVT hypotheses and setting up the equation for c

Consider the function f(x)=e−x2f(x) = e^{-x^2} on the closed interval [−1,1][-1, 1].

The function f(x)=e−x2f(x) = e^{-x^2} is continuous on [−1,1][-1,1] and differentiable on (−1,1)(-1,1), so the Mean Value Theorem guarantees at least one number cc in (−1,1)(-1,1) such that f′(c)=f(1)−f(−1)1−(−1)f'(c) = \dfrac{f(1)-f(-1)}{1-(-1)}. Since ff is even, the right-hand side equals 00. Solve the resulting equation for cc numerically (you may use graphing or numerical root-finding), and give an approximate value of cc in (−1,1)(-1,1) satisfying the MVT conclusion.

Hint

First compute f′(x)=−2xe−x2f'(x) = -2xe^{-x^2}. Since f(1)=f(−1)f(1) = f(-1), the secant slope is 0, so you need to solve f′(c)=0f'(c) = 0 for cc in (−1,1)(-1,1).

Step-by-step solution

Since f(x)=e−x2f(x)=e^{-x^2} is a composition of continuous, differentiable elementary functions, it is continuous on [−1,1][-1,1] and differentiable on (−1,1)(-1,1), so the MVT applies. Compute f(1)=e−1f(1) = e^{-1} and f(−1)=e−1f(-1) = e^{-1}, so the average rate of change is f(1)−f(−1)2=0\dfrac{f(1)-f(-1)}{2} = 0.

We need f′(c)=0f'(c) = 0 where f′(x)=−2xe−x2f'(x) = -2xe^{-x^2} (by the chain rule). Setting −2ce−c2=0-2ce^{-c^2}=0 and noting e−c2e^{-c^2} is never zero, we require c=0c=0. This value lies in (−1,1)(-1,1), confirming the MVT guarantee.

Verifying MVT hypotheses and setting up the equation for c — solution graph

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap