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

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- Nov 11th 2012, 11:27 PMcamjensonLooking 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.

- Nov 11th 2012, 11:36 PMMarkFLRe: Looking for inflection points
Have you found the second derivative?

- Nov 11th 2012, 11:38 PMcamjensonRe: Looking for inflection points
Yes, it's e^x(cos(e^x)-e^xsin(e^x))

- Nov 12th 2012, 12:02 AMMarkFLRe: Looking for inflection points
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. - Nov 12th 2012, 12:08 AMcamjensonRe: Looking for inflection points
I'm sorry. I don't know Newton's method?

- Nov 12th 2012, 12:11 AMSorobanRe: Looking for inflection points
Hello, camjenson!

Your second derivative is incorrect.

Quote:

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: .

- Nov 12th 2012, 12:17 AMcamjensonRe: Looking for inflection points
WHat do you mean by back-substitute x=ln(u)? Sorry I'm a little slow...

- Nov 12th 2012, 12:20 AMMarkFLRe: Looking for inflection points
- Nov 12th 2012, 11:05 PMhollywoodRe: Looking for inflection points
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