Thread: Functions to find Values

I've tried every way I could think of, but I can't seem to get it right. How do you find the value of g(f(a)) when g(x)= x^2-2 and f(x)=-2x+7 p.s. I know the answer is 4a^2-28a+47..but I want to work through the problem.

You offer us and . Did you mean instead of , or vice versa?

I fixed it.

g(x) = x^2 – 2
f(x) = -2x+7
First substitute x as a
f(a) = -2a +7
then
g(f(a)) = g(-2a+7)
= (-2a+7)^2 - 2
= 4a^2 - 28a + 47

Originally Posted by Mathivadhana99

g(x) = x^2 – 2
f(x) = -2x+7
First substitute x as a
f(a) = -2a +7
then
g(f(a)) = g(-2a+7)
= (-2a+7)^2 - 2
= 4a^2 - 28a + 47

Excellent!

What formula did you use to get 4a^ 2-28a+4 from (-2a+7)^ 2-2?

you apply (a+b)^2 formula
(a+b)^2 = a^2 + 2ab + b^2

(-2a+7)^2 -2 = (-2a)^2 + 2(-2a)(7) + (7)^2 - 2
= 4a^2 - 28a + 49 - 2
= 4a^2 - 28a +47

the formula Mathivadhana99 used above is the same one i used:



.

since multiplication of real numbers is commutative:

. so, continuing, we get:

.

this is basic, and just uses the distributive law:

repeatedly.