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
1 answer:
Answer:
Below in bold.
Step-by-step 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.
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