Analysis of Functions: Calculus

Question

Find the intervals of upward and downward concavities and coordinates of the points of inflections:

f(x)=3x⁴-4x³+1

Answer 1

You find the take the second derivative first.

The first derivative is 12x³-12x².

The second derivative is 36x²-24x.

Now, set the second derivative equal to zero and solve.

36x²-24x=0

12x(3x-2)=0

x=0, 2/3

Then, we test values based on those answers. We have to test a value less than 0, a value between 0 and 2/3, and a value greater than 2/3. We will test -1, 0.5, and 1.

Plug these values into the second derivative.

x=-1: 12(-1)(3(-1)-2) =-12(-5)=60 POSITIVE

x=0.5: 12(0.5)(3(0.5)-2)= 6(-0.5)=-3 NEGATIVE

x=1: 12(1)(3(1)-2) = 12(1)=12 POSITIVE

The intervals will be as such:

Concavity upward: (-∞, 0) and (2/3, ∞)

Concavity downward: (0, 2/3)

Since we alternate between positive and negative values above, we know the points of inflection are x=0 and x=2/3. Plug these x-values into f(x) to get the y-coordinate. That means we have (0, 1) and (2/3, 11/27).