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