# Functions to find Values

• Jan 16th 2013, 05:30 PM
Yommz
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.
• Jan 16th 2013, 05:37 PM
abender
Re: 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 PM
Yommz
Re: Functions to find Values
I fixed it.
• Jan 16th 2013, 06:12 PM
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
• Jan 16th 2013, 06:29 PM
abender
Re: Functions to find Values
Quote:

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!
• Jan 16th 2013, 08:46 PM
Yommz
Re: 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 PM
Deveno
Re: 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 AM
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
• Jan 17th 2013, 10:27 AM
Deveno
Re: 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.