# How do you find the relative extrema for #f(x) = x^2(6-x)^3#?

##### 1 Answer

Minimum when

#### Explanation:

Find the critical values of the function. These occur when the derivative equals

To find the derivative of the function, first find the derivative. Although you could distribute the equation, it's probably easier to use the product rule.

#f'(x)=(6-x)^3d/dx[x^2]+x^2d/dx[(6-x)^3]#

Find each derivative (the second requires the chain rule):

#d/dx[x^2]=2x#

#d/dx[(6-x)^3]=3(6-x)^2*d/dx[6-x]=-3(6-x)^2#

Plug these back in.

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

Factor

#f'(x)=x(6-x)^2(2(6-x)-3x)#

Simplify.

#f'(x)=x(6-x)^2(12-5x)#

This is never undefined. It is equal to

We can determine what types of extrema these are using the first derivative test (see how the signs change around the points).

Determining

When

#x<0# ,#f'(x)<0# .

When#0 < x <12/5,f'(x)>0#

Since the derivative changes from decreasing to increasing, there is a minimum at#x=0# .

Determining

When

#0 < x <12/5,f'(x)>0#

When#12/5 < x < 6,f'(x)<0#

Since the derivative changes from increasing to decreasing, there is a maximum at#x=12/5# .

Determining

When

#12/5 < x < 6,f'(x)<0#

When#x>6,f'(x)<0#

Here, the sign of the second derivative doesn't change. This means it is not an extremum, just a point where graph temporarily flattens.

graph{x^2(6-x)^3 [-3, 9, -100, 300]}