# What are the points of inflection, if any, of f(x)=x^(1/3)*(x+3)^(2/3) ?

Nov 11, 2016

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

$f ' \left(x\right) = {\left(x\right)}^{\frac{1}{3}} \left(\frac{2}{3}\right) {\left(x + 3\right)}^{- \frac{1}{3}} + \left(\frac{1}{3}\right) {\left(x\right)}^{- \frac{2}{3}} {\left(x + 3\right)}^{\frac{2}{3}}$
$f ' \left(x\right) = \frac{2 {x}^{\frac{1}{3}}}{3 {\left(x + 3\right)}^{\frac{1}{3}}} + {\left(x + 3\right)}^{\frac{2}{3}} / \left(3 {x}^{\frac{2}{3}}\right)$
$f ' \left(x\right) = \frac{6 x + 3 \left(x + 3\right)}{9 \left(x + 3\right)}$
$f ' \left(x\right) = \frac{9 \left(x + 1\right)}{9 \left(x + 3\right)}$
$f ' \left(x\right) = \frac{x + 1}{x + 3}$

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

No points of inflection of f(x) because you cannot set numerator of the second derivative of f(x) equal to zero.