# How do you find the maximum, minimum and inflection points and concavity for the function f(x) = (4x)/(x^2+4)?

Sep 25, 2016

$\textcolor{red}{x = 2}$
$\textcolor{red}{x = - 2}$

#### Explanation:

Given -

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

Find the first Derivative.

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

${f}^{'} \left(x\right) = \frac{4 {x}^{2} + 16 - 8 {x}^{2}}{{x}^{2} + 4} ^ 2$

${f}^{'} \left(x\right) = \frac{- 4 {x}^{2} + 16}{{x}^{2} + 4} ^ 2$

Find the Second Derivative

${f}^{'} ' \left(x\right) = \frac{\left[{\left({x}^{2} + 4\right)}^{2} \left(- 8 x\right)\right] - \left[\left(- 4 {x}^{2} + 16\right) \left(2\right) \left({x}^{2} + 4\right) \left(2 x\right)\right]}{{\left({x}^{2} + 4\right)}^{2}} ^ 2$

${f}^{'} ' \left(x\right) = \frac{\left[{\left({x}^{2} + 4\right)}^{2} \left(- 8 x\right)\right] - \left[4 x \left(- 4 {x}^{2} + 16\right) \left({x}^{2} + 4\right)\right]}{{x}^{2} + 4} ^ 4$

${f}^{'} ' \left(x\right) = \frac{\left({x}^{2} + 4\right) \left[\left({x}^{2} + 4\right) \left(- 8 x\right)\right] - \left[4 x \left(- 4 {x}^{2} + 16\right)\right]}{{x}^{2} + 4} ^ 4$

${f}^{'} ' \left(x\right) = \frac{\left[\left({x}^{2} + 4\right) \left(- 8 x\right)\right] - \left[4 x \left(- 4 {x}^{2} + 16\right)\right]}{{x}^{2} + 4} ^ 3$

f^''(x)= (-8x^3-32x+16x^3-64x)/(x^2+4)^3

${f}^{'} ' \left(x\right) = \frac{8 {x}^{3} - 96 x}{{x}^{2} + 4} ^ 3$

To find the Maxima and Minima, set the 1st derivative equal to zero.

${f}^{'} \left(x\right) = 0 \implies \frac{- 4 {x}^{2} + 16}{{x}^{2} + 4} ^ 2 = 0$

$- 4 {x}^{2} + 16 = 0$
${x}^{2} = \frac{- 16}{- 4} = 4$
$x = \pm \sqrt{4}$

$\textcolor{red}{x = 2}$
$\textcolor{red}{x = - 2}$

$x$ has two values

At $x = 2$

${f}^{'} ' \left(x\right) = \frac{8 {x}^{3} - 96 x}{{x}^{2} + 4} ^ 3 = \frac{8 {\left(2\right)}^{3} - 96 \left(2\right)}{{\left(2\right)}^{2} + 4} ^ 3 = \frac{64 - 192}{1728} = \frac{- 128}{1728} < 0$

At x=2, f^'(x)=0; f^''<0

Hence the function has a maximum at $x = 2$

At $x = - 2$

${f}^{'} ' \left(x\right) = \frac{8 {x}^{3} - 96 x}{{x}^{2} + 4} ^ 3 = \frac{8 {\left(- 2\right)}^{3} - 96 \left(- 2\right)}{{\left(- 2\right)}^{2} + 4} ^ 3 = \frac{64 + 192}{1728} = \frac{256}{1728} < 0$

At x=2, f^'(x)=0; f^''>0

Hence the function has a minimum at $x = - 2$