For $g (x) = x^{3} + 2 x^{2} + c x + k$ $g(x)=x^(3) +2x^(2) +cx +k$ (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 $y = g (x) and y = g^{- 1} (x)$ $y=g(x) and y=g^(-1) (x)$ , find the values of k?

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Answer 1 · 2018-05-13T05:28:51

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

a) We know that stationary points occur when $\frac{d y}{d x} = 0$ $dy/dx = 0$ .

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

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.

enter image source here

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

Hopefully this helps!