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

Question

Samantha Adkison · Accepted Answer

Critical points at #x = 0#

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!

Evelyn Agard · Answer

undefined To find the critical values of $ f(x) = \frac{x^3}{\sqrt{x + 25}} $, we need to first find its derivative and then solve for $ x $ when the derivative equals zero.

First, let's find the derivative of $ f(x) $ using the quotient rule:

$$ f'(x) = \frac{(3x^2)(\sqrt{x + 25}) - x^3 \left(\frac{1}{2\sqrt{x + 25}}\right)}{(x + 25)} $$

Simplify the expression:

$$ f'(x) = \frac{3x^2\sqrt{x + 25} - \frac{x^3}{2\sqrt{x + 25}}}{x + 25} $$

Now, set $ f'(x) $ equal to zero and solve for $ x $:

$$ 3x^2\sqrt{x + 25} - \frac{x^3}{2\sqrt{x + 25}} = 0 $$

Multiply both sides by $ 2\sqrt{x + 25} $ to eliminate the fraction:

$$ 6x^2(x + 25) - x^3 = 0 $$

Expand and rearrange terms:

$$ 6x^3 + 150x^2 - x^3 = 0 $$
$$ 5x^3 + 150x^2 = 0 $$
$$ x^2(5x + 150) = 0 $$

Now, solve for $ x $:

$$ x^2 = 0 \implies x = 0 $$

or

$$ 5x + 150 = 0 $$
$$ 5x = -150 $$
$$ x = -30 $$

So, the critical values of $ f(x) $ are $ x = 0 $ and $ x = -30 $.