How do use the first derivative test to determine the local extrema #f(x)= 4x^3 - 3x^4#?

Question

Scarlett Allison · Accepted Answer

The local extrema occur at point where the first derivative is equal to zero (critical points) and where the slope (as given by the first derivative) switches between positive and negative

Given #f(x) = 4x^3-3x^4# #(d(f(x)))/(dx) = 12x^2-12x^3# For critical points #color(white)("XXXX")12x^2-12x^3=0# #rarrcolor(white)("XX")x*x*12*(1-x)=0# #rarrcolor(white)("XX")x=0 and x=1# are critical points Consider three points: #color(white)("XXX")x=-1# (below the smallest critical point) #color(white)("XXX")x=1/2# (between the critical points) #color(white)("XXX")x=2# (above the largest critical point) #(df)/(dx)(-1) = 12*(-1)^2 - 12*(-1)^3 = 24# #(df)/(dx)(1/2) = 12*(1/2)^2-12*(1/2)^3 =12/8# #(df)/(dx)(2) = 12*(2)^2-12*(2)^3 = -48# The slope remains positive on each side of #x=0# so #x=0# gives a point of inflection which is not a local extrema. The slope changes from positive to negative on the two sides of #x=1# so #f(1)=1# is a local maximum. The situation can be further examined by considering the graph below: graph{4x^3-3x^4 [-2.733, 2.744, -1.366, 1.374]}

Paisley Johnson · Answer

undefined To use the first derivative test to determine the local extrema of $ f(x) = 4x^3 - 3x^4 $:

1. Find the first derivative of the function, $ f'(x) $.
2. Set $ f'(x) = 0 $ and solve for $ x $ to find critical points.
3. Determine the sign of $ f'(x) $ around each critical point.
4. If $ f'(x) $ changes sign from positive to negative at a critical point, it indicates a local maximum. If $ f'(x) $ changes sign from negative to positive, it indicates a local minimum.

Let's go through the steps:

1. $ f(x) = 4x^3 - 3x^4 $
   
   $ f'(x) = 12x^2 - 12x^3 $

2. Set $ f'(x) = 0 $:

$ 12x^2 - 12x^3 = 0 $

$ 12x^2(1 - x) = 0 $

$ x = 0 $ or $ x = 1 $

These are the critical points.

3. Test the sign of $ f'(x) $ around each critical point:

Test $ x = -1 $: $ f'(-1) = 12(-1)^2 - 12(-1)^3 = 12 - 12 = 0 $ (neither positive nor negative)
   
   Test $ x = 0 $: $ f'(0) = 12(0)^2 - 12(0)^3 = 0 $ (neither positive nor negative)
   
   Test $ x = 1 $: $ f'(1) = 12(1)^2 - 12(1)^3 = 12 - 12 = 0 $ (neither positive nor negative)

4. Since the sign of $ f'(x) $ doesn't change around the critical points, there are no local extrema for this function.