# How do you find the maximum, minimum and inflection points and concavity for the function y=16x + 1/x^2?

Jan 8, 2018

See below.

#### Explanation:

First we need to find the first derivative and solve this for $0$. This will identify any stationary points, because max/min points have a gradient of $0$. Inflection points can have a gradient of $0$ as well, or they can have undefined gradients. ( perpendicular ).

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(16 x + {x}^{- 2}\right) = 16 - 2 {x}^{- 3} = 16 - \frac{2}{x} ^ 3$

$f ' \left(x\right) = 0$

$16 - \frac{2}{x} ^ 3 = 0$

$\frac{2}{x} ^ 3 = 16$

$\frac{1}{x} ^ 3 = 8 \implies x = \sqrt{\frac{1}{8}} = \frac{1}{2}$ for real $x$

Using second derivative, if:

$f ' ' \left(x\right) < 0$ maximum point

$f ' ' \left(x\right) > 0$ minimum point

$f ' ' \left(x\right) = 0$ minimum/maximum or point of inflection

$f ' ' \left(x\right) = f ' \left(f ' \left(x\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(16 - \frac{2}{x} ^ 3\right) = 6 {x}^{- 4} = \frac{6}{x} ^ 4$

Plugging in $\frac{1}{2}$

$\frac{6}{\frac{1}{2}} ^ 4 > 0$

Minimum point at $\left(\frac{1}{2} , 12\right)$

For concavity, if:

$f ' ' \left(x\right) > 0$ convex ( concave up )

$f ' ' \left(x\right) < 0$ concave ( concave down )

$\frac{6}{x} ^ 4 > 0$

$\left(- \infty , 0\right) \cup \left(0 , \infty\right)$

Convex $\left(- \infty , 0\right) \cup \left(0 , \infty\right)$

No inflection point, vertical asymptote at $x = 0$

$\frac{6}{x} ^ 4 < 0$ ( no real solutions )

GRAPH: 