Question

Let f(x) = x^3 + px^2 + qx ?

a) Find the values of p and q so that f(-1) = -8 and f'(-1) = 12

b) Find the value of p so that graph of f changes concavity at x = 2

c) Under what conditions of p and q will the graph of f be increasing everywhere?

Answer 1

a) For the first condition to be true, `-8 = (-1)^3 + p(-1)^2 + q(-1) -> -8 = -1 + p - q -> -7 = p - q`

The derivative with respect to `x` is `f'(x) = 3x^2 + 2px + q`. For condition two to be valid, `12 = 3(-1)^2 + 2(-1)p + q -> 12 = 3 - 2p + q -> 9 = q - 2p`

So we have a system of equations.

`{(p - q = -7), (q - 2p = 9):}`

Solving through elimination we get `-p = 2 -> p = -2` which means that `q = 5`.

b) Concavity is determined by the second derivative.

`f'(x) = 3x^2 + 2px + q`
`f''(x) = 6x + 2p`

For `f(x)` to change concavity `f''(x) = 0`.

`0 = 6x + 2p`

`0 = 6(2) + 2p`

`-12 = 2p`

`p = -6`

c) For a function to be increasing on all `x`, `f'(x) > 0` on all `x`.

`f'(x) = 3x^2 + 2px + q`

`0 < 3x^2 + 2px + q`

I don't know what you can do to simplify this further.

Hopefully this helps!