using chain rule to find tangent to f(x) = x^2(x-5)^3

[smileykd]

Find f'(x), and find the value(s) of x where the tangent line is horizontal.

f(x) = x^2(x-5)^3

Using the chain rule, I know the first step is:
2x(x-5)^3+3x^2(x-5)^2

My textbook says the next step is:
5x(x-5)^2(x-2)
and i understand the 5x(x-5)^2 part, but i don't know how they got the (x-2).

can someone explain?

 

[galactus]

Here's how they got that:

\(\displaystyle \L\\x^{2}(x-5)^{3}\)

Product rule:

\(\displaystyle \L\\x^{2}(3)(x-5)^{2}+2x(x-5)^{3}(2x)\)

\(\displaystyle \L\\3x^{2}(x-5)^{2}+2x(x-5)^{3}\)

Factor out \(\displaystyle x(x-5)^{2}\):

\(\displaystyle \L\\x(x-5)^{2}[3x+2(x-5)]\)

\(\displaystyle \L\\x(x-5)^{2}[5x-10]\)

Factor out 5:

\(\displaystyle \H\\5x(x-5)^{2}(x-2)\)

 

[soroban]

Re: using chain rule

Hello, smileykd!

Find \(\displaystyle f'(x)\), and find the value(s) of x where the tangent line is horizontal.
. . \(\displaystyle f(x) \:= \:x^2(x\,-\,5)^3\)

Using the chain rule, I know the first step is:
. . \(\displaystyle f'(x) \:= \:2x(x\,-\,5)^3\,+\,3x^2(x\,-\,5)^2\;\) **

My textbook says the next step is: \(\displaystyle \,5x(x\,-\,5)^2(x\,-\,2)\)

and i understand the \(\displaystyle 5x(x\,-\,5)^2\) part, . . . You do?
but i don't know how they got the \(\displaystyle (x\,-\,2).\)

**
At this point, we are going to solve: \(\displaystyle \,2x(x\,-\,5)^3\,+\,3x^2(x\,-\,5)^2\:=\:0\) . . . right?

So we factor: \(\displaystyle \,x(x\,-\,5)^2\,\left[2(x\,-\,5)\,+\,3x\right] \:=\:0\;\;\Rightarrow\;\;x(x\,-\,5)^2\,\left[2x\,-\,10\,+\,3x\right] \:=\:0\)

. . \(\displaystyle x(x\,-\,5)^2[5x\,-\,10]\:=\:0\;\;\Rightarrow\;\;x(x\,-\,5)^2\cdot5(x\,-\,2)\:=\:0\)

And we have: \(\displaystyle \L\,5x(x\,-\,5)^2(x\,-\,2)\;\) . . . Got it?

Edit: Beat me again, Cody!

 

[smileykd]

Thanks