# If f(x) = (5/2)x^(2/3) - x^(5/3), what are the points of inflection, concavity and critical points?

Nov 7, 2016

$f \left(x\right) = \left(\frac{5}{2}\right) {x}^{\frac{2}{3}} - {x}^{\frac{5}{3}}$

The domain of $f$ is $\left(- \infty , \infty\right)$.

$f ' \left(x\right) = \left(\frac{5}{3}\right) {x}^{- \frac{1}{3}} - \frac{5}{3} {x}^{\frac{2}{3}}$

$= \frac{5}{3} \left[{x}^{- \frac{1}{3}} - {x}^{\frac{2}{3}}\right]$

$= \frac{5}{3} \left[\frac{1 - x}{x} ^ \left(\frac{1}{3}\right)\right]$

$f '$ does not exist at $x = 0$ and $f ' \left(x\right) = 0$ at $x = 1$.
Both $0$ and $1$ are in the domain of $f$, so both are critical numbers.

$f ' ' \left(x\right) = \frac{5}{3} \left[- \frac{1}{3} {x}^{- \frac{4}{3}} - \frac{2}{3} {x}^{- \frac{1}{3}}\right]$

$= - \frac{5}{9} {x}^{- \frac{4}{3}} \left[1 + 2 x\right]$

$= - \frac{5 \left(1 + 2 x\right)}{9 {x}^{\frac{4}{3}}}$

$f ' ' \left(x\right) > 0$ for $x < - \frac{1}{2}$ and

$f ' ' \left(x\right) > 0$ for $- \frac{1}{2} < x < 0$ and $x > 0$

The graph of $f$ is concave up on $\left(- \infty , - \frac{1}{2}\right)$ and concave down on $\left(- \frac{1}{2} , 0\right)$ and on $\left(0 , \infty\right)$.

The point $\left(- \frac{1}{2} , f \left(- \frac{1}{2}\right)\right)$ is an inflection point.