Thread: Looking for inflection points

Find the first positive x value of x for which f(x)=sin(e^x) has an inflection point. Correctly take derivatives.

Have you found the second derivative?

Yes, it's e^x(cos(e^x)-e^xsin(e^x))

Correct. Since for all real , this leaves you to solve:



I recommend a numeric root-finding technique, such as Newton's method. Graph the function to determine a good initial guess.

I'm sorry. I don't know Newton's method?

Hello, camjenson!

Your second derivative is incorrect.

Find the first positive value of for which has an inflection point.

We have: .

Then: .

Hence: .

And we must solve: .


Equation [1] has no solutions.


Equation [2] can be approximated.

Let
We have: .
And we can approximate a root of: .
Then back-substitute: .

WHat do you mean by back-substitute x=ln(u)? Sorry I'm a little slow...

Originally Posted by Soroban

...

Your second derivative is incorrect.

The OP has the same result that you cite.

Originally Posted by camjenson

I'm sorry. I don't know Newton's method?

You should be able to find it in your textbook, or online. I assume you are expected to know it, since you have been given this problem.

Or perhaps the professor expects you to use a graphing calculator to find the root. The calculator uses Newton's method, but you don't need to know that to plug in the functions and get your answer.

- Hollywood