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