What x values is the function concave down if f(x) = 15x^(2/3) + 5x?

Feb 25, 2015

$f \left(x\right) = 15 {x}^{\frac{2}{3}} + 5 x$ is concave downward for all $x < 0$

As Kim suggested a graph should make this apparent (See bottom of this post).

Alternately,

Note that $f \left(0\right) = 0$
and checking for critical points by taking the derivative and setting to $0$

we get
$f ' \left(x\right) = 10 {x}^{- \frac{1}{3}} + 5 = 0$
or
$\frac{10}{x} ^ \left(\frac{1}{3}\right) = - 5$

which simplifies (if $x < > 0$) to
${x}^{\frac{1}{3}} = - 2$
$\rightarrow$ $x = - 8$

At $x = - 8$
$f \left(- 8\right) = 15 {\left(- 8\right)}^{\frac{2}{3}} + 5 \left(- 8\right)$
$= 15 {\left(- 2\right)}^{2} + \left(- 40\right)$
$= 20$

Since ($- 8 , 20$) is the only critical point (other than ($0 , 0$) )
and $f \left(x\right)$ decreases from $x = - 8$ to $x = 0$

it follows that $f \left(x\right)$ decreases on each side of ($- 8 , 20$), so

$f \left(x\right)$ is concave downward when $x < 0$.

When $x > 0$ we simply note that
$g \left(x\right) = 5 x$ is a straight line and
$f \left(x\right) = 15 {x}^{\frac{2}{3}} + 5 x$ remains a positive amount (namely $15 {x}^{\frac{2}{3}}$ above that line
therefore $f \left(x\right)$ is not concave downward for $x > 0$.

graph{15x^(2/3) + 5x [-52, 52, -26, 26]}