# How do you find intercepts, extrema, points of inflections, asymptotes and graph g(x)=x+32/x^2?

Dec 1, 2017

Please see below for a partial solution.

#### Explanation:

$g \left(x\right) = x + \frac{32}{x} ^ 2$

x intercepts
$g \left(x\right) = 0$ at $x = - \frac{32}{x} ^ 2$
which happens at ${x}^{3} = - 32$

so $x = \sqrt[3]{- 32} = - 2 \sqrt[3]{4}$

y intercept
None. $g \left(0\right)$ does not exist.

Asymptotes

${\lim}_{x \rightarrow 0} g \left(x\right) = \infty$, so $x = 0$ (the $y$-axis) is a verticle asymptote..

${\lim}_{x \rightarrow 00} g \left(x\right) = \infty$ so there is no horizontal asymptote.

${\lim}_{x \rightarrow \infty} \left(g \left(x\right) - x\right) = 0$ so $y = x$ is an oblique (slant) asymptote)

Analysis of first derivative

$g ' \left(x\right) = 1 - \frac{64}{x} ^ 3 = \frac{{x}^{3} - 64}{x} ^ 3$ is undefined at $x = 0$ and is $0$ at $x = 4$.

On $\left(- \infty , 0\right)$, we have $g ' \left(x\right) > 0$ so $g$ is increasing.
$x = 0$ is not a critical number.

On $\left(0 , 4\right)$, we have $g ' \left(x\right) < 0$ so $g$ is decreasing.
On $\left(4 , \infty\right)$, we have $g ' \left(x\right) > 0$ so $g$ is increasing.

$f \left(4\right) = 6$ is a local minimum.