# 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?

##### 1 Answer

#### Explanation:

a) We know that stationary points occur when

#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

#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

#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!