 Nikhil Holt

2022-03-14

What x values is the function concave down if
$f\left(x\right)=15{x}^{\frac{2}{3}}+5x$? orangepaperiz7

Step 1
$f\left(x\right)=15{x}^{\frac{2}{3}}+5x$ 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
${f}^{\prime }\left(x\right)=10{x}^{-\frac{1}{3}}+5=0$
or
$\frac{10}{{x}^{\frac{1}{3}}}=-5$
which simplifies (if $x<>0$) to
${x}^{\frac{1}{3}=-2}$
$\to x=-8$
At $x=-8$
$f\left(-8\right)=15{\left(-8\right)}^{\frac{2}{3}}+5\left(-8\right)$
$=15{\left(-2\right)}^{2}+\left(-40\right)$
Since  is the only critical point (other than  ) and f(x) decreases from $x=-8$ to $x=0$
it follows that $f\left(x\right)$ decreses on each side of , so
$f\left(x\right)$ is concave downward when $x<0$
When $x>0$ we simplify note that
$g\left(x\right)=5x$ is a straight line and
$f\left(x\right)=15{x}^{\frac{2}{3}}+5x$ remains a positive amount (namely $15{x}^{\frac{2}{3}}$ above that line therefore f(x) iss not concave downward for $x>0$

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