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Chapter 4: Problem 65

Determine the intervals on which the following functions are concave up or concave down. Identify any inflection points. $$f(x)=e^{-x^{2} / 2}$$

Short Answer

Expert verified
Question: Determine the intervals on which the function \(f(x) = e^{-x^2 / 2}\) is concave up and concave down, and find any inflection points. Solution: The function is concave up on the interval \((-1, 1)\), concave down on the intervals \((-\infty, -1)\) and \((1, \infty)\), and has inflection points at \((-1, e^{-1/2})\) and \((1, e^{-1/2})\).

Step by step solution

01

Find the first and second derivatives

Given the function \(f(x) = e^{-x^2 / 2}\), we first determine the first derivative \(f'(x)\), and then the second derivative \(f''(x)\). Using the chain rule, we have: $$f'(x) = -x e^{-x^2 / 2}$$ To find the second derivative, again use the chain rule as well as the product rule: $$f''(x) = e^{-x^2 / 2} - x^2 e^{-x^2 / 2}$$.
02

Determine concave up and concave down intervals

A function is concave up when its second derivative is positive and concave down when its second derivative is negative. To find these intervals, we need to determine the signs of \(f''(x)\). Factor out the common term \(e^{-x^2 / 2}\): $$f''(x) = e^{-x^2 / 2}(1 - x^2)$$ The exponential term \(e^{-x^2 / 2}\) is always positive. Therefore, the sign of \(f''(x)\) is determined by the sign of \((1 - x^2)\). - \(f''(x) > 0\) (concave up) for \(-1 < x < 1\) - \(f''(x) < 0\) (concave down) for \(x < -1\) or \(x > 1\)
03

Identify inflection points

Inflection points occur when the second derivative changes its sign. From our analysis in Step 2, we see that the sign change of \(f''(x)\) occurs at \(x = -1\) and \(x = 1\). So, we have two inflection points at: $$x = -1: f(-1) = e^{-1^2 / 2} = e^{-1/2}$$ $$x = 1: f(1) = e^{-1^2 / 2} = e^{-1/2}$$ In conclusion: - The function is concave up on the interval \((-1, 1)\) - The function is concave down on the intervals \((-\infty, -1)\) and \((1, \infty)\) - There are inflection points at \((-1, e^{-1/2})\) and \((1, e^{-1/2})\).

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