Question

determine the interval(s) on which the function (f(x)=\frac{18}{x^{5}}) is concave down.
concave down:

Explanation:

Step1: Find the first - derivative

Given $f(x)=\frac{18}{x^{5}} = 18x^{-5}$. Using the power - rule $(x^n)^\prime=nx^{n - 1}$, we have $f^\prime(x)=18\times(-5)x^{-6}=-90x^{-6}$.

Step2: Find the second - derivative

Differentiate $f^\prime(x)$ with respect to $x$. Using the power - rule again, $f^{\prime\prime}(x)=(-90)\times(-6)x^{-7}=540x^{-7}=\frac{540}{x^{7}}$.

Step3: Determine where $f^{\prime\prime}(x)<0$

A function $y = f(x)$ is concave down when $f^{\prime\prime}(x)<0$. Set $\frac{540}{x^{7}}<0$. Since $540>0$, then $x^{7}<0$. Solving $x^{7}<0$ gives $x < 0$.

Answer:

$(-\infty,0)$