# How do you find the local max and min for f(x) = x / (x^2 + 81)?

Sep 20, 2016

$f \left(9\right) = \frac{1}{18} \text{ is local Maxima.}$

$f \left(- 9\right) = - \frac{1}{18} \text{ is local Minima.}$

#### Explanation:

We know that $f$ has a local maxima /minima at $x = a$, then,

(1) f'(a)=0, and, (2) f''(a) lt or gt 0" according as maxima or minima, resp."

Now, $f \left(x\right) = \frac{x}{{x}^{2} + 81} \therefore f ' \left(x\right) = \frac{\left({x}^{2} + 81\right) \left(1\right) - x \left(2 x\right)}{{x}^{2} + 81} ^ 2 , i . e . ,$

$f ' \left(x\right) = \frac{81 - {x}^{2}}{{x}^{2} + 81} ^ 2$.

$\therefore f ' \left(x\right) = 0 \Rightarrow x = \pm 9$

Further, $f ' ' \left(x\right) = \frac{{\left({x}^{2} + 81\right)}^{2} \left(- 2 x\right) - \left(81 - {x}^{2}\right) 2 \left({x}^{2} + 81\right) 2 x}{{x}^{2} + 81} ^ 4$

$= \frac{\left({x}^{2} + 81\right) \left(- 2 x\right) \left({x}^{2} + 81 + 2 \left(81 - {x}^{2}\right)\right)}{{x}^{2} + 81} ^ 4$

$= \frac{- 2 x \left(3 \left(81\right) - {x}^{2}\right)}{{x}^{2} + 81} ^ 3$

Now, $f ' ' \left(9\right) = \frac{- 2 \left(9\right) \left(3 \left(81\right) - 81\right)}{81 + 81} ^ 3 = \frac{- 2 \left(9\right) \left(2 \left(81\right)\right)}{{2}^{3} \cdot {81}^{3}} < 0$

$\Rightarrow f \left(9\right) = \frac{9}{81 + 81} = \frac{9}{2 \cdot 81} = \frac{1}{18} \text{ is local Maxima.}$

Finally, $f ' ' \left(- 9\right) = \frac{- 2 \left(- 9\right) \left(2 \left(81\right)\right)}{{2}^{3} \cdot {81}^{3}} > 0$

$\Rightarrow f \left(- 9\right) = - \frac{9}{2 \cdot 81} = - \frac{1}{18} \text{ is local Minima.}$

Sep 20, 2016

$\text{The Minima is "-1/18", and, the Maxima is } \frac{1}{18.}$

#### Explanation:

Let $x = 9 \tan \theta , \theta \in \mathbb{R} - \left\{\left(2 n + 1\right) \frac{\pi}{2} : n \in \mathbb{Z}\right\}$

Then, $f \left(x\right) = \frac{9 \tan \theta}{81 {\tan}^{2} \theta + 81} = \frac{9 \tan \theta}{81 {\sec}^{2} \theta}$

$= \frac{1}{9.} \sin \frac{\theta}{\cos} \theta \cdot {\cos}^{2} \theta = \frac{1}{9} \sin \theta \cos \theta$, or,

$= \frac{1}{18} \left(2 \sin \theta \cos \theta\right) = \frac{1}{18} \sin 2 \theta$

But, we know that,

$\forall \theta \in \mathbb{R} - \left\{\left(2 n + 1\right) \frac{\pi}{2} : n \in \mathbb{Z}\right\} , - 1 \le \sin 2 \theta \le 1$.

Clearly, $\text{the Minima is "-1/18", and, the Maxima is } \frac{1}{18.}$