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

Beginner2022-03-15Added 9 answers

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 $x=-8$

$f(-8)=15{(-8)}^{\frac{2}{3}}+5(-8)$

$=15{(-2)}^{2}+(-40)$

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

it follows that $f\left(x\right)$ decreses on each side of $(-8,\text{}20)$, 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 $x>0$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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