# Math Help - Functions to find Values

1. ## 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.

2. ## Re: Functions to find Values

You offer us $g(x)$ and $h(x)$. Did you mean $f(x)$ instead of $h(x)$, or vice versa?

I fixed it.

4. ## Re: Functions to find Values

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

5. ## Re: Functions to find Values

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!

6. ## Re: Functions to find Values

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

7. ## Re: Functions to find Values

$(-2a+7)^2 - 2 = (-2a+7)(-2a+7) - 2 = (-2a)(-2a + 7) + (7)(-2a+7) - 2$

$= (-2a)(-2a) + (-2a)(7) + (7)(-2a) + (7)(7) - 2$

$= 4a^2 - 14a - 14a + 49 - 2 = 4a^2 - 28a + 47$

8. ## Re: Functions to find Values

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

9. ## Re: Functions to find Values

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

$(a+b)^2 = (a+b)(a+b) = (a)(a+b) + (b)(a+b)$

$= (a)(a) + (a)(b) + (b)(a) + (b)(b)$.

since multiplication of real numbers is commutative:

$(b)(a) = ba = ab = (a)(b)$. so, continuing, we get:

$= a^2 + ab + ba + b^2 = a^2 + ab + ab + b^2 = a^2 + (ab + ab) + b^2 = a^2 + 2ab + b^2$.

this is basic, and just uses the distributive law:

$a(b+c) = ab + ac; (a+b)c = ac + bc$ repeatedly.