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

**JOhkonut** · August 28th 2008, 04:40 PM

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

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

**Jhevon** · August 28th 2008, 05:19 PM

Originally Posted by JOhkonut

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

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: $f'(x) = 3x^2 \bigg( \ln \frac {\pi}3\bigg) \bigg( \frac {\pi}3\bigg)^{x^3 - 8}$

and

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

**JOhkonut** · August 28th 2008, 05:25 PM

Originally Posted by Jhevon

hint: $f'(x) = 3x^2 \bigg( \ln \frac {\pi}3\bigg) \bigg( \frac {\pi}3\bigg)^{x^3 - 8}$

and

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

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...

**JOhkonut** · August 28th 2008, 05:28 PM

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

**Jhevon** · August 28th 2008, 05:35 PM

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-- $3x^2$ is a function as well as $\bigg( \frac {\pi}3 \bigg)^{x^3 - 8}$ , hence we need the product rule, which yields two terms)

**JOhkonut** · August 28th 2008, 05:37 PM

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

This makes things much easier, thanks!

**11rdc11** · August 28th 2008, 05:46 PM

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

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

oops too slow again sorry the latex slowed me down

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