SOLUTION: f (x) = x^2 + x and g (x)= x - 5. a. Find h (x) = (f infinity g)(x). b. State the domain of h (x) = (f infinity g) (x). c. Find h (x) = (g infinity f)(x) d. State the domain of h (
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-> SOLUTION: f (x) = x^2 + x and g (x)= x - 5. a. Find h (x) = (f infinity g)(x). b. State the domain of h (x) = (f infinity g) (x). c. Find h (x) = (g infinity f)(x) d. State the domain of h (
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Question 238818: f (x) = x^2 + x and g (x)= x - 5. a. Find h (x) = (f infinity g)(x). b. State the domain of h (x) = (f infinity g) (x). c. Find h (x) = (g infinity f)(x) d. State the domain of h (x)= (g infinity f) (x). Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! f (x) = x^2 + x and g (x)= x - 5.
Comment: That symbol is not infinity.
It means "compose" or "form the composite".
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a. Find h (x) = (f o g)(x).
fog(x) = f[g(x)] = f[x-5] = (x-5)^2+(x-5) = x^2 -26x +20
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b. State the domain of h (x) = (f o g) (x).
Domain:
The range of g(x) is all Real Numbers
so the domain of h(x) is all Real Numbers.
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c. Find h (x) = (g o f)(x)
gof(x) = g[f(x)] = g[x^2+x] = (x^2+x)-5 = x^2+x-5
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d. State the domain of h (x)= (g o f) (x).
The Range of f(x) = ?
f(x)=x^2+x
a = 1, b = 1
Vertex is at -b/2a = -1/2
f(-1/2) = (-1/2)^2+(-1/2) = 1/4 - 1/2 = -1/4
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So Range of f(x) is all Rean Numbers >= -1/4
So the Domain of h(x) is all Real Numbers greater than or equal to -1/4.
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Cheers,
Stan H.
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