# How do you find the inflection points for the function f(x)=e^sin(x)?

Oct 2, 2014

Inflection points are when a function changes concavity.

A simple way to think of concavity is by asking yourself if a function seems more like a smile or a frown at a certain point. A "smiling" function is concave up, and a "frowning" function is considered to be concave down.

So a point of inflection is where a function changes from smiling to frowning.

Inflection points are found with the 2nd derivative. Wherever the 2nd derivative is positive, you have a smiling function. Wherever the 2nd derivative is negative, you have a frowning function.

So the point of inflection is going to be where the change from smiling to frowning, or smiling to frowning occurs.

This is often, but not always at a point where the 2nd derivative equals 0. Finding the zeros of the 2nd derivative is a good start, but you are really interested in the sign changes of the 2nd derivative.

Given $f \left(x\right) = {e}^{\sin} \left(x\right)$, we need to find $f ' ' \left(x\right)$.

$f ' \left(x\right) = {e}^{\sin} \left(x\right) \cos \left(x\right)$

by The Chain Rule.

Finding the second derivative will require The Product and The Chain Rule. Using the Chain Rule as:

$d \left[u v\right] = u ' v + u v '$, and $u = {e}^{\sin} \left(x\right)$ and $v = \cos \left(x\right)$, we get:

$f ' ' \left(x\right) = \left[{e}^{\sin} \left(x\right) \cos \left(x\right) \left(\cos \left(x\right)\right) + {e}^{\sin} \left(x\right) \left(- \sin \left(x\right)\right)\right]$

$f ' ' \left(x\right) = {e}^{\sin} \left(x\right) {\cos}^{2} \left(x\right) - \sin x {e}^{\sin} \left(x\right)$

$f ' ' \left(x\right) = {e}^{\sin} \left(x\right) \left[{\cos}^{2} \left(x\right) - \sin x\right]$

We need to find where this function crosses the x-axis (when $f ' ' \left(x\right)$ changes from negative to positive or vice versa) in order to find the inflection points of $f \left(x\right)$.

If you have access to technology for this question, then you can simply graph it. But you might not be allowed to use technology, and the function can be confusing visually.

My preference is look at each factor separately.

$f ' ' \left(x\right) = {e}^{\sin} \left(x\right) \left[{\cos}^{2} \left(x\right) - \sin x\right]$ consists of

${e}^{\sin} \left(x\right)$

and

$\left[{\cos}^{2} \left(x\right) - \sin x\right]$.

${e}^{\sin} \left(x\right)$ will never be zero or negative. So we can ignore it for purposes of determining the sign of the 2nd derivative (which will tell us the concavity of the original function.

$\left[{\cos}^{2} \left(x\right) - \sin x\right]$ is more promising. Basically, you will have a zero of the 2nd derivative every time ${\cos}^{2} \left(x\right) - \sin \left(x\right) = 0$

I will admit that solving this equation without technology is daunting. So I turned to WolphramAlpha.com and input "second derivative of e^(sinx)". The result was:

The infomation under "Roots above tell you how to find the zeros of the 2nd derivative for integer values of n. And the graphs under "Plots" show that these roots are all points of inflection, because the 2nd derivative changes sign at each x-intercept.

Hope this helps. This one had a particularly answer in the end.