Question

The function h(x) = x^3+bx+d has a critical point at (2,-4). Determine the constants b and d and find the equation of h(x). Show your work.

Answer 1

well, you know that critical points occur when h'(x)=0. So, we want

3x^2+b = 0 at x=2.
So, b = -12

That means that

h(x) = x^3-12x+d
h(2) = 8-24+d = -4
So, d = 12, giving us

h(x) = x^3-12x+12

The graph is at

http://www.wolframalpha.com/input/?i=x^3-12x%2B12+for+x%3D0..3

Looks like there's a minimum at (2,-4) all right.

Answer 2

h ' (x) = 3x^2 + b
= 0 at a critical point
if (2,-4) is a critical point,
3(4) + b = 0
b = -12
but (2,-4) also satisfies h(x)
-4 = 8 - 12(2) + c
c = 12

thus h(x) = x^3 - 12x + 12