# How many points of inflection does the function f(x) = (pi/3)^((x^3)-8) have?

Mar 29, 2015

One.

To find the points of inflection of any function, we compute the function's derivative then find the points where the derivative equals zero. In this case, we use the chain rule: let $u = {x}^{3} - 8$, then

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} {\left(\frac{\pi}{3}\right)}^{{x}^{3} - 8} = \left(\frac{d}{\mathrm{dx}} {\left(\frac{\pi}{3}\right)}^{u}\right) \left(\frac{d}{\mathrm{dx}} \left({x}^{3} - 8\right)\right)$
$\implies f ' \left(x\right) = {\left(\frac{\pi}{3}\right)}^{{x}^{3} - 8} \ln \left(\frac{\pi}{3}\right) \left(3 {x}^{2}\right)$

There is no need to simplify this function any further. We know that for $f ' \left(x\right)$ to equal zero, then at least one of three things must happen:

${\left(\frac{\pi}{3}\right)}^{{x}^{3} - 8} = 0$, or
$\ln \left(\frac{\pi}{3}\right) = 0$, or
$3 {x}^{2} = 0$.

Clearly, $\ln \left(\frac{\pi}{3}\right)$ is not zero. Nor is ${\left(\frac{\pi}{3}\right)}^{{x}^{3} - 8}$, as any function $f \left(x\right) = {a}^{p \left(x\right)}$ is never zero for any polynomial $p \left(x\right)$. This leaves $3 {x}^{2}$, and we know that $3 {x}^{2} = 0$ only when $x = 0$. Hence the one and only point of inflection is $x = 0$.

Aug 6, 2015

Two points of inflections:
x=0 , f(0) = (pi/3)^(-8) " and " x=-2/ln(pi/3), f(-2/ln(pi/3)) = (pi/3)^(-8(1/(ln(pi/3))^3+1)

#### Explanation:

A point of inflection can be obtained when $f ' ' \left(x\right) = 0$.

Using the chain rule: let $u = {x}^{3} - 8$, then

$f \left(x\right) = {\left(\frac{\pi}{3}\right)}^{{x}^{3} - 8}$
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} {\left(\frac{\pi}{3}\right)}^{{x}^{3} - 8} = \frac{d}{\mathrm{du}} {\left(\frac{\pi}{3}\right)}^{u} \frac{\mathrm{du}}{\mathrm{dx}}$
$= {\left(\frac{\pi}{3}\right)}^{{x}^{3} - 8} \ln \left(\frac{\pi}{3}\right) 3 {x}^{2}$

$f ' ' \left(x\right) = 3 \ln \left(\frac{\pi}{3}\right) {\left(\frac{\pi}{3}\right)}^{{x}^{3} - 8} x \left(x \ln \left(\frac{\pi}{3}\right) + 2\right) = 0$

Since:
${\left(\frac{\pi}{3}\right)}^{{x}^{3} - 8} > 0$
only possible solution is when:
$x \left(x \ln \left(\frac{\pi}{3}\right) + 2\right) = 0$.

Hence two points of inflections:
x=0 , f(0) = (pi/3)^(-8) " and " x=-2/ln(pi/3), f(-2/ln(pi/3)) = (pi/3)^(-8(1/(ln(pi/3))^3+1)

Note that when $x = 0 , f \left(0\right)$ it is also a stationary point.