# What are the critical values, if any, of f(x)= x/(x^2 + 25)?

Jan 19, 2018

Required Critical Points are: color(blue)(x=+5, x=-5

#### Explanation:

Given:

$\textcolor{red}{f \left(x\right) = \frac{x}{{x}^{2} + 25}}$

Obviously, the domain of this function is: color(blue)((-oo < x < +oo) and our function is defined.

Critical Points are points where the function is defined and its derivative is zero or undefined

$\textcolor{g r e e n}{S t e p .1}$

We have,

$\textcolor{b l u e}{y = f \left(x\right) = \frac{x}{{x}^{2} + 25}}$

We will now differentiate our $f \left(x\right)$

i.e., find $\frac{d}{\mathrm{dx}} \left[\frac{x}{{x}^{2} + 25}\right]$

Quotient Rule is used to differentiate.

Quotient Rule is given by

$\textcolor{b l u e}{{\left[\frac{u \left(x\right)}{v \left(x\right)}\right]}^{'} = \frac{u ' \left(x\right) . v \left(x\right) - u \left(x\right) \cdot v ' \left(x\right)}{v {\left(x\right)}^{2}}}$

$\frac{d}{\mathrm{dx}} \left[\frac{x}{{x}^{2} + 25}\right]$

$\Rightarrow \frac{\frac{d}{\mathrm{dx}} \left(x\right) \cdot \left({x}^{2} + 25\right) - x \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + 25\right)}{{x}^{2} + 25} ^ 2$

$\Rightarrow \frac{1 \cdot \left({x}^{2} + 25\right) - \left\{\frac{d}{\mathrm{dx}} \left({x}^{2}\right) + \frac{d}{\mathrm{dx}} \left(25\right)\right\} \cdot x}{{x}^{2} + 25} ^ 2$

$\Rightarrow \frac{{x}^{2} - 2 {x}^{2} + 25}{{x}^{2} + 25} ^ 2$

$\Rightarrow \frac{- {x}^{2} + 25}{{x}^{2} + 25} ^ 2$

Hence,

$\textcolor{b r o w n}{f ' \left(x\right) = \frac{- {x}^{2} + 25}{{x}^{2} + 25} ^ 2}$

$\textcolor{g r e e n}{S t e p .2}$

Set

color(brown)(f'(x) = 0

Hence,

$\textcolor{b r o w n}{f ' \left(x\right) = \frac{- {x}^{2} + 25}{{x}^{2} + 25} ^ 2 = 0}$

For a rational function, the derivative will be equal to zero, if the expression in the numerator is equal to zero

Set,

$- {x}^{2} + 25 = 0$

Add $\textcolor{red}{- 25}$ to both sides of the equation to get

-x^2 + 25+ color(red)((-25)) = 0+ color(red)((-25)

-x^2 + cancel 25+ color(red)((- cancel 25)) = 0+ color(red)((-25)

$- {x}^{2} = - 25$

Divide both sides by color(red)((-1)

(-x^2)/color(red)((-1)) =-25/color(red)((-1)

${x}^{2} = 25$

$\therefore x = \pm 5$

$x = + \mathmr{and} x = - 5$ are also on the domain of our function

Required Critical Points are: color(blue)(x=+5, x=-5