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.

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- Jan 16th 2013, 05:30 PMYommzFunctions 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.

- Jan 16th 2013, 05:37 PMabenderRe: Functions to find Values
You offer us $\displaystyle g(x)$ and $\displaystyle h(x)$. Did you mean $\displaystyle f(x)$ instead of $\displaystyle h(x)$, or vice versa?

- Jan 16th 2013, 05:57 PMYommzRe: Functions to find Values
I fixed it.

- Jan 16th 2013, 06:12 PMMathivadhana99Re: 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 - Jan 16th 2013, 06:29 PMabenderRe: Functions to find Values
- Jan 16th 2013, 08:46 PMYommzRe: Functions to find Values
What formula did you use to get 4a^ 2-28a+4 from (-2a+7)^ 2-2?

- Jan 16th 2013, 09:19 PMDevenoRe: Functions to find Values
$\displaystyle (-2a+7)^2 - 2 = (-2a+7)(-2a+7) - 2 = (-2a)(-2a + 7) + (7)(-2a+7) - 2$

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

$\displaystyle = 4a^2 - 14a - 14a + 49 - 2 = 4a^2 - 28a + 47$ - Jan 17th 2013, 07:44 AMMathivadhana99Re: 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 - Jan 17th 2013, 10:27 AMDevenoRe: Functions to find Values
the formula Mathivadhana99 used above is the same one i used:

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

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

since multiplication of real numbers is commutative:

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

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

this is basic, and just uses the distributive law:

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