How do you find concavity when f(x)= x^(7/3) + x^(4/3)?

Dec 6, 2015

To find concavity, use the second derivative of the function.

If $f ' ' \left(a\right) < 0$, the function is concave down (also called concave) when $x = a$.

If $f ' ' \left(a\right) > 0$, the function is concave up (also called convex) when $x = a$.

If $f ' ' \left(a\right) = 0$, the point when $x = a$ is when the concavity of a function shifts called the point of inflection.

$f ' \left(x\right) = \frac{7}{3} {x}^{\frac{4}{3}} + \frac{4}{3} {x}^{\frac{1}{3}}$

$f ' ' \left(x\right) = \frac{28}{9} {x}^{\frac{1}{3}} + \frac{4}{9} {x}^{- \frac{2}{3}}$

$f ' ' \left(x\right) = \frac{28 x + 4}{9 {x}^{\frac{2}{3}}}$

graph{(28x+4)/(9x^(2/3)) [-20.28, 20.27, -10.14, 10.14]}

This is a graph of $f ' ' \left(x\right)$.

Notice how the $f ' ' \left(x\right) = 0$ when $x = - \frac{1}{7}$. This means that when $x = - \frac{1}{7}$ there is a point of inflection on $f \left(x\right)$.

$f ' ' \left(x\right) < 0$ when $x < - \frac{1}{7}$, so $f \left(x\right)$ is concave down whenever $x < - \frac{1}{7}$.

Similarly, $f \left(x\right)$ is concave up whenever $x > - \frac{1}{7}$.