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

Question

Paisley Johnson · Accepted Answer

Minimum when #x=0#, maximum when #x=12/5#

Find the critical values of the function. These occur when the derivative equals #0# or is undefined.

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 #x(6-x)^2# from each term.

#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 #0# when #x=0#, #x=6#, or #x=12/5#.

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

Determining #x=0#:

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 #x=12/5#:

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 #x=6#:

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]}

Ezekiel Alexander · Answer

undefined To find the relative extrema of $ f(x) = x^2(6-x)^3 $, we first find the critical points by setting the derivative equal to zero and solving for $ x $. Then, we use the second derivative test to determine if these critical points correspond to relative minima, relative maxima, or neither.

1. Find the derivative of $ f(x) $:
   $ f'(x) = 2x(6-x)^3 + x^2 \cdot 3(6-x)^2 \cdot (-1) $
   $ f'(x) = 2x(6-x)^2[3(6-x) - x] $
   $ f'(x) = 2x(6-x)^2(18-3x - x) $
   $ f'(x) = 2x(6-x)^2(18-4x) $
   $ f'(x) = 2x(6-x)^2(18-4x) $

2. Set the derivative equal to zero and solve for $ x $:
   $ 2x(6-x)^2(18-4x) = 0 $
   $ x(6-x)^2(18-4x) = 0 $
   $ x = 0, x = 6 $ (multiplicity 2)

3. Find the second derivative $ f''(x) $:
   $ f''(x) = 2(6-x)^2(18-4x) + 2x(6-x)^2(-4) + 2x(18-4x) \cdot 2(6-x) $
   $ f''(x) = 2(6-x)^2[18-4x - 4x] + 2x(6-x)^2(-4) $
   $ f''(x) = 2(6-x)^2(18-8x) - 8x(6-x)^2 $
   $ f''(x) = 2(6-x)^2(18-8x) - 8x(6-x)^2 $
   $ f''(x) = 2(6-x)^2(18-8x) - 8x(6-x)^2 $

4. Evaluate $ f''(0) $, $ f''(6) $, and $ f''(3) $:
   $ f''(0) = 2(6-0)^2(18-8 \cdot 0) - 8 \cdot 0(6-0)^2 = 2 \cdot 6^2 \cdot 18 = 648 $
   $ f''(6) = 2(6-6)^2(18-8 \cdot 6) - 8 \cdot 6(6-6)^2 = 0 $
   $ f''(3) = 2(6-3)^2(18-8 \cdot 3) - 8 \cdot 3(6-3)^2 = 2 \cdot 3^2 \cdot 2 = 36 $

5. Analyze the results:
   - At $ x = 0 $ and $ x = 6 $, the second derivative test is inconclusive because $ f''(0) $ and $ f''(6) $ are both zero.
   - At $ x = 3 $, since $ f''(3) = 36 $ (positive), the function has a relative minimum at $ x = 3 $.