# Thread: Looking for inflection points

1. ## 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.

2. ## Re: Looking for inflection points

Have you found the second derivative?

3. ## Re: Looking for inflection points

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

4. ## Re: Looking for inflection points

Correct. Since $\displaystyle 0<e^x$ for all real $\displaystyle x$, this leaves you to solve:

$\displaystyle f(x)=\cos(e^x)-e^x\sin(e^x)=0$

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

5. ## Re: Looking for inflection points

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

6. ## Re: Looking for inflection points

Hello, camjenson!

Find the first positive value of $\displaystyle x$ for which $\displaystyle f(x)\:=\:\sin(e^x)$ has an inflection point.

We have: .$\displaystyle f(x) \:=\:\sin(e^x)$

Then: .$\displaystyle f'(x) \:=\:e^x\cdot\cos(e^x)$

Hence: .$\displaystyle f''(x) \;=\; e^x\!\cdot\!\cos(e^x) + e^x\!\cdot\![-\sin(e^x)]\!\cdot\! e^x \;=\;e^x\big[\cos(e^x) - e^x\sin(e^x)\big]$

And we must solve: .$\displaystyle \begin{Bmatrix}e^x \:=\:0 & [1] \\ \cos(e^x) - e^x\sin(e^x) \:=\:0 & [2] \end{Bmatrix}$

Equation [1] has no solutions.

Equation [2] can be approximated.

Let $\displaystyle u = e^x$
We have: .$\displaystyle \cos u - u\sin u \:=\:0 \quad\Rightarrow\quad u\sin u \:=\:\cos u \quad\Rightarrow\quad \frac{\sin u}{\cos u} \:=\:\frac{1}{u}$
And we can approximate a root of: .$\displaystyle \tan u \:=\:\frac{1}{u}$
Then back-substitute: .$\displaystyle x \,=\,\ln(u)$

7. ## Re: Looking for inflection points

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

8. ## Re: Looking for inflection points

Originally Posted by Soroban
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