Question

For `g(x)=x^(3) +2x^(2) +cx +k` (c and k are real numbers) : a. If g has exactly one stationary point find the value of c. b. If the stationary point occurs at a point of intersection of `y=g(x) and y=g^(-1) (x)`, find the values of k?

Answer 1

`c = 4/3`
`k = -10/27`

Explanation:

a) We know that stationary points occur when `dy/dx = 0`.

`g'(x) = 3x^2 +4x +c`

By the discriminant,

`b^2 - 4ac = 0` for there to be only one solution to `g'(x) = 0`.

`4^2 - 4(3)(c) = 0`

`16 = 12c`

`4/3 = c`

b) Let's start by finding the value of `x` where the stationary point is

`0 = 3x^2 + 4x + 4/3`

`0 = 9x^2 + 12x + 4`

`0 = 9x^2 + 6x + 6x + 4`

`0 = 3x(3x+ 2) + 2(3x + 2)`

`0 = (3x+ 2)^2`

`x = -2/3`

Now recall that a function and its inverse will always intersect on the line `y = x`. Thus, the solution point will be

`x = x^3 + 2x^2 + cx + k`

Therefore

`-2/3 = (-2/3)^3 + 2(-2/3)^2 + (4/3)(-2/3) + k`

`-2/3 = -8/27 + 8/9 - 8/9 + k`

`-2/3 + 8/27 = k`

`-10/27 = k`

We can confirm graphically.

As you can see, the lower point of intersection is indeed a stationary point on the red graph (horizontal tangent).

Hopefully this helps!