SOLUTION: Please help me solve this. I think this is subtraction of functions. Here is the problem: f(a)= a^3+5a^2-2a+4 g(a)=-3a^2+3a^3+2a-4, solve for (g-f) (a)

Question 1187530: Please help me solve this. I think this is subtraction of functions. Here is the problem:
f(a)= a^3+5a^2-2a+4
g(a)=-3a^2+3a^3+2a-4, solve for (g-f) (a)
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
(g-f)(a) is equal to g(a) - f(a).
g(a) = -3a^2+3a^3+2a-4
f(a) = a^3+5a^2-2a+4
you subtract like terms from like terms.
3a^3 - a^3 = 2a^3
-3a^2 - 5a^2 = -8a^2
2a - -2a = 4a
-4 - 4 = -8
your resultant equation is (g-f)(a) = 2a^3 - 8a^2 + 4a - 8

to see if you did it correctly, assign a random value to a.
i used 7.

(g-f)(a) = 2a^3 - 8a^2 + 4a - 8 = 314 when a = 7.

g(a) = -3a^2 + 3a^3 + 2a - 4 = 892 when a = 7.

f(a) = a^3 + 5a^2 - 2a + 4 = 578 when a = 7.

892 - 578 = 314.

you get the same answer when you let a = 7.
(g-f)(a) = 314
g(a) = 892
f(a) = 578
g(a) - f(a) = 314.

this confirms that (g-f)(a) = g(a) - f(a) was evaluated correctly.

RELATED QUESTIONS
This question is on the Addition Method rather than the Substitution Method. When using... (answered by Abbey,guapa)

Can you help me with this problem? 3a^2-2a plus 7)plus a^2 -3a-4) my plus sign is not... (answered by tutor_paul)

I need to find the quotient using long division. 9a^3-3a^2-3a+4 divided by 3a+2 I... (answered by richwmiller)

Your astute algebra knowledge and assistance is needed to help me solve this problem.... (answered by chiexpert)

Is this problem factored completely correctly? 3a^3+6a^2-2a-4 (3a^3)+(-2a-4)... (answered by scott8148)

i need help with this problem. 3a^3+10a^2-5a+12 divided by... (answered by stanbon,Gogonati)

Would you please verify that I have correctly done the below problem? 3a/a+4 - 2a^2 -... (answered by Edwin McCravy,MathLover1)

Please help me solve this equation: 2a+7/a^2-2a-15 -... (answered by Fombitz)

I have been working on this problem for quite a wile but I cant seem to figure it out... (answered by ankor@dixie-net.com)