Question

The functions it f and g are defined for all real values of x by f(x)=x^2+4ax+a^2 and g(x)=4x-2a where a is a positive constant. Given that fg(3)=69 find the value of a and hence find the value of x such that g^-1(x) =x

Answer 1

Below in bold.

Explanation:

We find fg(x) by replacing the x in f(x) by g(x), so:

fg(x) = (4x - 2a)^2 + 4a(4x - 2a) + a^2

fg(3) = 89 so:

(4(3) - 2a)^2 + 4a(4(3) - 2a) + a^2 = 69

(12 - 2a)^2 + 48a - 8a^2 + a^2 - 69 = 0

144 - 48a + 4a^2 + 48a - 8a^2 + a^2 - 69 = 0

-3a^2 + 75 = 0

-3a^2 = -75

a^2 = 25

a = 5.

Now g(x) = 4x - 2(5) = 4x - 10.

Finding g-1(x);

4x = g(x) + 10

x = (g(x) + 10)/4

So g-1(x) = (x + 10)/4

When g-1(x)= x

x = (x +10)/4

4x = x + 10

3x = 10

x = 10/3.