SOLUTION: Let f(x)=x^2+3x-5,g(x)=x-3, find a. (f+g)(x) b. (f-g)(x) c. (fg)(x) d. (f/g)(x) e.(f of g)(x) f. g[f(x)] g.(f-g)(-2) h. f[f(3)] Please help!!!

Question 58693: Let f(x)=x^2+3x-5,g(x)=x-3, find
a. (f+g)(x)
b. (f-g)(x)
c. (fg)(x)
d. (f/g)(x)
e.(f of g)(x)
f. g[f(x)]
g.(f-g)(-2)
h. f[f(3)]
Please help!!!
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
Let f(x)=x^2+3x-5,g(x)=x-3, find
a. (f+g)(x)=(x^2+3x-5)+(x-3)
(f+g)(x)=x^2+3x+x-5-3

:
b. (f-g)(x)=(x^2+3x-5)-(x-3)
(f-g)(x)=x^2+3x-5-x+3
(f-g)(x)=x^2+3x-x-5+3

:
c. (fg)(x)=(x^2+3x-5)(x-3)
(fg)(x)=x(x^2+3x-5)-3(x^2+3x-5)
(fg)(x)=x^3+3x^2-5x-3x^2-9x+15
(fg)(x)=x^3+3x^2-3x^2-5x-9x+15

:
d. (f/g)(x)=(x^2+3x-5)/(x-3) Nothing cancels by factoring so you're done.

:
e.(f of g)(x)=f(g(x))=f(x-3) substitute x-3 in for x in f:
(fog)(x)=(x-3)^2+3(x-3)-5
(fog)(x)=x^2-6x+9+3x-9-5
(fog)(x)=x^2-6x+3x+9-9-5

:
f. g[f(x)]=g(x^2+3x-5) substitute x^2+3x-5 in for x in g
g(f(x))=(x^2+3x-5)-3

:
g.(f-g)(-2) Take your answer from b and substitute -2 in for x.
(f-g)(x)=x^2+2x-2
(f-g)(-2)=(-2)^2+2(-2)-2
(f-g)(-2)=4-4-2

:
h. f[f(3)]
Find f(3), f(3)=(3)^2+3(3)-5=9+9-5=13
f(f(3))=f(13)
f(f(3))=(13)^2+3(13)-5
f(f(3))=169+39-5

:
Happy Calculating!!!
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