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Simple function questions - checking answer. |
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| May16-11, 10:19 PM | #1 |
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Simple function questions - checking answer.
1. The problem statement, all variables and given/known data
1) Given f(x)=x+3, x∈R and g(x)=x2,x∈R,0=<x=<4, find the range of composite function fg(x). 2) Given f(x)=x+1,x∈R,x>=5 and g(x)=x1/2, x∈R,x>=0. Find the domain of fg(x). 2. Relevant equations 3. The attempt at a solution For 1), the answer 1 get is y=>3 , but the answer given is3=<y=<19. 2) the answer i get is x>=0 , but yet the answer provided is x>=25. So what answers do you guys get? Anything wrong with the answers provided by the answer sheet? Cause i couldn't find anything wrong with my solutions... Thanks in advance. |
| May16-11, 10:48 PM | #2 |
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You almost had #1. You just didn't handle the other part of the inequality!
If you write out your work we can find where you went wrong. |
| May17-11, 04:54 AM | #3 |
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Recognitions:
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If you restrict the domain of f(g(x)) to be x ≥ 25, then the range of g(x) would be g(x) ≥ 5, which "matches" the domain of f(x). So x ≥ 25 is right. |
| May17-11, 07:22 AM | #4 |
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Simple function questions - checking answer. |
| May17-11, 09:38 AM | #5 |
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Mentor
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You are given that the domain of f is restricted to x >= 5. So although there are many values of x that are in the domain of g, there are some values of g(x) that aren't in the domain of f. Some examples that use selected values of x. x = 0. g(0) = sqrt(0) = 0. Can we evaluate f(0)? No, because 0 is not >= 5. x = 1. g(1) = sqrt(1) = 1. Can we evaluate f(1)? No, because 1 is not >= 5. x = 4. g(4) = sqrt(4) = 2. Can we evaluate f(2)? No, because 2 is not >= 5. x = 9. g(9) = sqrt(9) = 3. Can we evaluate f(3)? No, because 3 is not >= 5. Do you see where I'm going with this? |
| May17-11, 01:57 PM | #6 |
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Mentor
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Since f(x) is defined only for x ≥ 5, find the values of x for which g(x) ≥ 5. |
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