- #1

- 438

- 0

i found gf(x)=x^2-2

is it true that the range of gf(x) is the same as the range of g(x) ? If so,

the range of g(x) is [-5 , infinity) and the range of gf(x) is [-2 , infinity)

why arent they the same ?

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- Thread starter thereddevils
- Start date

- #1

- 438

- 0

i found gf(x)=x^2-2

is it true that the range of gf(x) is the same as the range of g(x) ? If so,

the range of g(x) is [-5 , infinity) and the range of gf(x) is [-2 , infinity)

why arent they the same ?

- #2

- 28

- 0

I think everything is correct, why would you think the range of g(f(x)) is the same as the range of g(x)?

- #3

- 954

- 117

I'll pose the reverse question: why should they be the same? g(x) and gf(x) are different functions.

- #4

- 438

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Take a look at the diagram i attached . Isn't that the image of gf(x) the same as the image function g(x) ?

- #5

- 330

- 3

The range of [itex]f[/itex] needn't be the whole of the domain of [itex]g[/itex] so some of the images under [itex]g[/itex] may not occur in [itex]gf[/itex]. You only have [itex]range(gf)\subseteq range(g)[/itex]. In fact you could say they're not equal because you have yourself provided a counterexample.

- #6

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thanks Martin , how about the domains , is the domain of f(x) the same as domain gf(x) because both of them started from the same set or it needn't also be the same in this case.

- #7

- 330

- 3

Yes, [itex]dom(gf)=dom(f)[/itex], assuming you only

- #8

- 438

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definethe composition [itex]gf[/itex] when [itex]range(f)\subseteq dom(g)[/itex].

thanks again Martin !

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