Question

Let $f(x) = x^2 + 4x - 31$. for what value of $a$ is there exactly one real value of $x$ such that $f(x) = a$?

Answer 1

To solve for this, we need to find for the value of x when the 1st derivative of the equation is equal to zero (or at the extrema point).

So what we have to do first is to derive the given equation:

f (x) = x^2 + 4 x – 31

 

Taking the first derivative f’ (x):

f’ (x) = 2 x + 4

 

Setting f’ (x) = 0 and find for x:

2 x + 4 = 0

x = - 2

 

Therefore the value of a is:

a = f (-2)

a = (-2)^2 + 4 (-2) – 31

a = 4 – 8 – 31

a = - 35