# Relative Max and Relative Min

• Oct 10th 2012, 11:51 PM
arcticreaver
Relative Max and Relative Min
f(x) = (x-2)^3 / x^2

find the relative max and min.

so obviously i would bring the x^2 up.

this is what i have so far,

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

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

​is this correct so far?
• Oct 10th 2012, 11:54 PM
MarkFL
Re: Relative Max and Relative Min
If you wish to use the product rule rather than the quotient rule by bringing the denominator up as a factor, then you need to negate its exponent.
• Oct 11th 2012, 12:11 AM
arcticreaver
Re: Relative Max and Relative Min
let me re-write this.

x^2(x-2)^3 - (x-2)^3-(x^2) / x^4

and the next step, i believe is the derivative right? but i am a bit confused with (x-2)^3. is the derivative of this 3(x-2)^2 or is it just 3x^2?
• Oct 11th 2012, 12:20 AM
MarkFL
Re: Relative Max and Relative Min
We are given the function:

$\displaystyle f(x)=\frac{(x-2)^3}{x^2}$

Using the quotient rule, we find:

$\displaystyle f'(x)=\frac{x^2(3(x-2)^2)-2x(x-2)^3}{(x^2)^2}=\frac{x(x-2)^2(3x-2(x-2))}{x^4}=\frac{(x-2)^2(x+4)}{x^3}$