Question

Let f(x) = e^(-1/2(x^2)) Which table describes the concavity of f?

Answer 1

Answer:

Concave Up: x<-1 and x>1

Concave Down: -1<x<1

Step-by-step explanation:

To determine the intervals of concavity for a continuous function f, we must determine the second derivative and figure out at what points its sign changes (aka. our inflection points):

$f(x)=e^{-\frac{1}{2}x^2}\\\\f'(x)=\frac{d}{dx}(-\frac{1}{2}x^2)*e^{-\frac{1}{2}x^2}\\\\f'(x)=-xe^{-\frac{1}{2}x^2}\\\\f''(x)=[\frac{d}{dx}(-x)](e^{-\frac{1}{2}x^2})+(-x)(\frac{d}{dx}e^{-\frac{1}{2}x^2})\\ \\f''(x)=-e^{-\frac{1}{2}x^2}+x^2e^{-\frac{1}{2}x^2}\\\\f''(x)=e^{-\frac{1}{2}x^2}(x^2-1)$

Next, we determine where $f''(x)=0$ to find our inflection points:

$0=-e^{-\frac{1}{2}x^2}(x^2-1)\\\\x=-1,x=1$

Our intervals of concavity can be determined by testing points around our inflection points. I will use the points $x=-2$ and $x=2$:

$f''(-2)=e^{-\frac{1}{2}(-2)^2}((-2)^2-1)=e^{-2}(4-1)=\frac{3}{e^2}>0\\ \\\\f''(2)=e^{-\frac{1}{2}(2)^2}(2^2-1)=e^{-2}(4-1)=\frac{3}{e^2}>0$

Since the factor $x^2-1$ is a second-order polynomial with a leading positive coefficient, we know that the sign is negative in the interval between the roots, and positive outside. Therefore:

Concave Downward: $(-\infty,-1) \cup(1,\infty)$

Concave Upward: $(-1,1)$

This means that the 4th option is correct. Refer to the graph for a visual.