# Thread: UGGGH! How many points of inflection does f(x) have?

1. ## UGGGH! How many points of inflection does f(x) have?

$\displaystyle f(x)=(\frac {\pi}{3})^{x^3-8}$

Don't laugh at me, but I tried to do this algebraically, which was TERRIBLE.

I need to show work, whatever that means.

a. none
b. one
c. two
c. three
e. Infinitely many

2. Originally Posted by JOhkonut
$\displaystyle f(x)=(\frac {\pi}{3})^{x^3-8}$

Don't laugh at me, but I tried to do this algebraically, which was TERRIBLE.

I need to show work, whatever that means.

a. none
b. one
c. two
c. three
e. Infinitely many
hint: $\displaystyle f'(x) = 3x^2 \bigg( \ln \frac {\pi}3\bigg) \bigg( \frac {\pi}3\bigg)^{x^3 - 8}$

and

$\displaystyle f''(x) = 6x \cdot \ln \frac {\pi}3 \cdot \bigg( \frac {\pi}3 \bigg)^{x^3 - 8} + (3x^2)^2 \bigg( \ln \frac {\pi}3 \bigg)^2 \bigg( \frac {\pi}3 \bigg)^{x^3 - 8}$

3. Originally Posted by Jhevon
hint: $\displaystyle f'(x) = 3x^2 \bigg( \ln \frac {\pi}3\bigg) \bigg( \frac {\pi}3\bigg)^{x^3 - 8}$

and

$\displaystyle f''(x) = 6x \cdot \ln \frac {\pi}3 \cdot \bigg( \frac {\pi}3 \bigg)^{x^3 - 8} + (3x^2)^2 \bigg( \ln \frac {\pi}3 \bigg)^2 \bigg( \frac {\pi}3 \bigg)^{x^3 - 8}$
I already tried that, but at the end, how can I be sure to get all the zeros with my graphing calculator? I can't find the roots algebraically either...

4. wait, thats not the same second derivative as I got... I don't see how you got that second term. Mine was more complicated.

5. Originally Posted by JOhkonut
wait, thats not the same second derivative as I got... I don't see how you got that second term. Mine was more complicated.
Oh? what did you get? (note that we have a product of functions-- $\displaystyle 3x^2$ is a function as well as $\displaystyle \bigg( \frac {\pi}3 \bigg)^{x^3 - 8}$, hence we need the product rule, which yields two terms)

6. oh i see now, I treated $\displaystyle \ln \frac {\pi}{3}$ as a term rather than a constant.

This makes things much easier, thanks!

7. $\displaystyle u = 3x^2$
$\displaystyle du = 6x$
$\displaystyle v =(\frac {\pi}{3})^{x^3-8}$
$\displaystyle dv = 3x^2 \bigg( \frac {\pi}3\bigg)^{x^3 - 8}$

$\displaystyle ln \frac{\pi}{3}[(du)(v) + (dv)(u)]$

oops too slow again sorry the latex slowed me down