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?

May 13, 2018

$c = \frac{4}{3}$
$k = - \frac{10}{27}$

Explanation:

a) We know that stationary points occur when $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$.

$g ' \left(x\right) = 3 {x}^{2} + 4 x + c$

By the discriminant,

${b}^{2} - 4 a c = 0$ for there to be only one solution to $g ' \left(x\right) = 0$.

${4}^{2} - 4 \left(3\right) \left(c\right) = 0$

$16 = 12 c$

$\frac{4}{3} = c$

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

$0 = 3 {x}^{2} + 4 x + \frac{4}{3}$

$0 = 9 {x}^{2} + 12 x + 4$

$0 = 9 {x}^{2} + 6 x + 6 x + 4$

$0 = 3 x \left(3 x + 2\right) + 2 \left(3 x + 2\right)$

$0 = {\left(3 x + 2\right)}^{2}$

$x = - \frac{2}{3}$

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

$x = {x}^{3} + 2 {x}^{2} + c x + k$

Therefore

$- \frac{2}{3} = {\left(- \frac{2}{3}\right)}^{3} + 2 {\left(- \frac{2}{3}\right)}^{2} + \left(\frac{4}{3}\right) \left(- \frac{2}{3}\right) + k$

$- \frac{2}{3} = - \frac{8}{27} + \frac{8}{9} - \frac{8}{9} + k$

$- \frac{2}{3} + \frac{8}{27} = k$

$- \frac{10}{27} = k$

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!