What are the critical values, if any, of #f(x)= x^3/sqrt(x + 25)#?

1 Answer
Jan 7, 2017

Answer:

Critical points at #x = 0#

Explanation:

First, differentiate. Let #f(x) = (g(x))/(h(x))#. Then #g(x) = x^3# and #h(x) = sqrt(x + 25)#.

We can differentiate #g(x)# using the power rule. #g'(x) = 3x^2#

We need the chain rule for #h(x)#. Let #y = sqrt(u) = u^(1/2)# and #u = x + 25#. Then #dy/(du) = 1/2u^(-1/2) = 1/(2u^(1/2)#. and #(du)/dx = 1#.

#h'(x) = 1/(2u^(1/2)) * 1 = 1/(2sqrt(x + 25))#

Use the quotient rule now.

#f'(x) = (g'(x) * h(x) - g(x) * h'(x))/(h(x))^2#

#f'(x) = (3x^2(sqrt(x + 25)) - x^3/(2sqrt(x + 25)))/(sqrt(x + 25))^2#

#f'(x) = ((6x^2(x + 25) - x^3)/(2sqrt(x + 25)))/(x +25)#

#f'(x) = (5x^3 + 150x^2)/(2(x+ 25)^(3/2))#

There will be critical points whenever the derivative equals #0# or is undefined. The derivative is undefined at #x = -25#. Set the derivative to #0# to find the other critical points.

#0 = (5x^3 + 150x^2)/(2(x + 25)^(3/2))#

#0 = 5x^3 + 150x^2#

#0 = 5x^2(x + 30)#

#x = 0 and -30#

So, there will be additional critical points at #x = 0# and #x= -30#.

However, to be a critical number, the function has to be defined at the given point. Therefore, #x = -30# and #x= -25# are not critical numbers. #x = 0# is the only critical number.

Hopefully this helps!