# How do you find the local max and min for f(x)= (x^2)/(x-2)^2?

Jul 7, 2018

local min is at $\left(0 , 0\right)$

There is no local maximum.

#### Explanation:

Given: $f \left(x\right) = {x}^{2} / {\left(x - 2\right)}^{2}$

Find the first derivative:

Use the quotient rule of differentiation: $\left(\frac{u}{v}\right) ' = \frac{v u ' - u v '}{v} ^ 2$

Let u = x^2; " "u' = 2x; " "v = (x-2)^2; " "v' = 2(x-2)^1(1)

$f ' \left(x\right) = \frac{{\left(x - 2\right)}^{2} \left(2 x\right) - 2 {x}^{2} \left(x - 2\right)}{x - 2} ^ 4$

$f ' \left(x\right) = \frac{\left(x - 2\right) \left[2 x \left(x - 2\right) - 2 {x}^{2}\right]}{x - 2} ^ 4$

$f ' \left(x\right) = \frac{2 {x}^{2} - 4 x - 2 {x}^{2}}{x - 2} ^ 3 = \frac{- 4 x}{x - 2} ^ 3$

Find the critical point(s). Let $f ' \left(x\right) = 0$:

$f ' \left(x\right) = \frac{- 4 x}{x - 2} ^ 3 = 0$

-4x = 0 * (x-2)^3 = 0$critical value: $x = 0$$f \left(0\right) = 0$critical point: $\left(0 , 0\right)$Use the first derivative test. Set up intervals and test values in the intervals to see if they are positive or negative: intervals: $\text{ "(-oo, 0), " "x = 0, " } \left(0 , \infty\right)$test value: $\text{ " x = -1 " "x = 0 " } x = 1$$f ' \left(x\right) : \text{ " < 0 " "= 0" } > 0$When the slope changes from negative, to zero, to positive, we have a local minimum. local min is at $\left(0 , 0\right)$There is no local maximum. Jul 7, 2018 There is a local min at $\left(0 , 0\right)$#### Explanation: The function is $f \left(x\right) = \frac{{x}^{2}}{x - 2} ^ 2$This function is a quotient of $2$derivable functions The derivative of a quotient is $\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{{v}^{2}}$$u = {x}^{2}$, $\implies$, $u ' = 2 x$$v = {\left(x - 2\right)}^{2}$, $\implies$, $u ' = 2 \left(x - 2\right)$Therefore, $f ' \left(x\right) = \frac{\left(2 x\right) {\left(x - 2\right)}^{2} - \left(x - 2\right) \left(2 {x}^{2}\right)}{x - 2} ^ 4$$= \frac{2 x \left(x - 2\right) - 2 {x}^{2}}{x - 2} ^ 3$$= \frac{2 {x}^{2} - 4 x - 2 {x}^{2}}{x - 2} ^ 3$$= - \frac{4 x}{x - 2} ^ 3$The critical points are when $f ' \left(x\right) = 0$That is $x = 0$Let's build a variation chart $\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$0$$\textcolor{w h i t e}{a a a a a a a a}$$2$$\textcolor{w h i t e}{a a a a a}$$+ \infty$$\textcolor{w h i t e}{a a a a}$$f ' \left(x\right)$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a}$↘$\textcolor{w h i t e}{a a}$color(white)(aaa)↗$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a}$↘# There is a local min at $\left(0 , 0\right)\$

graph{x^2/(x-2)^2 [-11.82, 13.49, -3.49, 9.17]}