How do you find all critical point and determine the min, max and inflection given #f(x)=x^4-4x^3+20#?

Question

Isabella Ackerman · Accepted Answer

Please see the explanation below

Calculate the first and second derivatives The function is #f(x)=x^4-4x^3+20# Calculate the first derivative #f'(x)=4x^3-12x^2# #f'(x)=0# #=>#, #4x^3-12x^2=0# #=>#, #4x^2(x-3)=0# The critical points are #x=0# and #x=3# Construct a variation chart #color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##0##color(white)(aaaaaa)##3##color(white)(aaaa)##+oo# #color(white)(aaaa)##f'(x)##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+# #color(white)(aaaa)##f(x)##color(white)(aaaaa)##↘##color(white)(aaaa)##↘##color(white)(aaaa)##↗# There is a local minimum at #(3, -7)# Calculate the second derivative #f''(x)=12x^2-24x# The points of inflections are when #f''(x)=0# #12x^2-24x=0# #=>#, #12x(x-2)=0# #=>#, #x=0# and #x=2# The inflection points are #(0, 20)# and #(2,4)# Build a variation chart to determine the concavities #color(white)(aaaa)##" Interval "##color(white)(aaaa)##(-oo, 0)##color(white)(aaaa)##(0,2)##color(white)(aaaa)##(2,+oo)# #color(white)(aaaa)##" sign f''(x)"##color(white)(aaaaaaa)##+##color(white)(aaaaaaa)##-##color(white)(aaaaaaaa)##+# #color(white)(aaaa)##" f(x)"##color(white)(aaaaaaaaaaaa)##uu##color(white)(aaaaaaa)##nn##color(white)(aaaaaaaa)##uu# graph{x^4-4x^3+20 [-32.73, 32.24, -5.85, 26.6]}

Scarlett Allison · Answer

undefined To find the critical points, we first take the derivative of the function f(x). Then, we set the derivative equal to zero and solve for x to find the critical points.

The derivative of f(x) = x^4 - 4x^3 + 20 is f'(x) = 4x^3 - 12x^2.

Setting f'(x) equal to zero and solving for x:

4x^3 - 12x^2 = 0
4x^2(x - 3) = 0

This gives us two critical points: x = 0 and x = 3.

To determine if these critical points correspond to a minimum, maximum, or point of inflection, we use the second derivative test.

The second derivative of f(x) is f''(x) = 12x^2 - 24x.

Evaluating the second derivative at each critical point:

At x = 0:
f''(0) = 12(0)^2 - 24(0) = 0

At x = 3:
f''(3) = 12(3)^2 - 24(3) = 108 - 72 = 36

Since f''(0) = 0, the second derivative test is inconclusive at x = 0.

At x = 3, since f''(3) = 36, which is greater than zero, it indicates a local minimum.

Therefore, the function has a local minimum at x = 3.

Since there is no change in concavity at x = 0 (as f''(0) = 0), it is an inflection point.