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

Question

Christopher Allers · Accepted Answer

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

Given:

#color(red)(f(x) = 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

#color(green)(Step.1)#

We have,

#color(blue)(y=f(x) = x/(x^2 + 25))#

We will now differentiate our #f(x)#

i.e., find #d/(dx) [x/(x^2 + 25)]#

Quotient Rule is used to differentiate.

Quotient Rule is given by

#color(blue)([(u(x))/(v(x))]^' = (u'(x).v(x) - u(x)*v'(x))/(v(x)^2))#

#d/(dx) [x/(x^2 + 25)]#

#rArr [d/(dx)(x)*(x^2+25) - x*d/(dx)(x^2+25)]/(x^2+25)^2#

#rArr [1*(x^2+25)-{d/(dx)(x^2)+d/(dx)(25)}*x]/(x^2+25)^2#

#rArr (x^2 - 2x^2+25)/(x^2+25)^2#

#rArr ( - x^2+25)/(x^2+25)^2#

Hence,

#color(brown)(f'(x) = ( - x^2+25)/(x^2+25)^2)#

#color(green)(Step.2)#

Set

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

Hence,

#color(brown)(f'(x) = ( - 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 #color(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#

#:. x = +-5#

#x = + and x = -5# are also on the domain of our function

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

Emma Aldridge · Answer

undefined To find the critical values of $ f(x) = \frac{x}{x^2 + 25} $, we first need to find the derivative of $ f(x) $ with respect to $ x $ and then solve for $ x $ when the derivative equals zero.

The derivative of $ f(x) $ with respect to $ x $ is:

$$ f'(x) = \frac{(x^2 + 25)(1) - (x)(2x)}{(x^2 + 25)^2} $$

$$ = \frac{x^2 + 25 - 2x^2}{(x^2 + 25)^2} $$

$$ = \frac{25 - x^2}{(x^2 + 25)^2} $$

To find the critical values, we set $ f'(x) $ equal to zero and solve for $ x $:

$$ \frac{25 - x^2}{(x^2 + 25)^2} = 0 $$

$$ 25 - x^2 = 0 $$

$$ x^2 = 25 $$

$$ x = \pm 5 $$

So, the critical values of $ f(x) $ are $ x = -5 $ and $ x = 5 $.