Simple question...which I can't seem to get!

[|ll|lll|]

Wouldn't dom f(g(x)) = dom g(x)?

[Andiio]

Wouldn't dom f(g(x)) = dom g(x)?

Nope, the answer is different!

[nacho]

its not R \ {-5/2} ?

[Andiio]

its not R \ {-5/2} ?

That's what I thought it was, but the answer says it's (-∞,0)\{-5/2}

I don't get why they only have R- as opposed to R, since you need g(x) to be an element of R\{0}... :\

[nacho]

seems like an error in solutiions

just realised that g(x) was restricted
so it should be the domain of g? as mentioned by the lines guy

[Andiio]

seems like an error in solutiions

Can anyone confirm this? I've been killing myself over this D:

And nacho, just curious, how did you work out your answer to get that? Just curious as to your approach

[b^3]

Using the calc by defining the functions with the appropriate domain gives the graph with domain (- infinty, -5/2)

[brightsky]

ran g(x): (-infty, 0) which is a subset of dom f(x)
so dom fog should be dom g(x)...which is (-infty, -5/2)

the answer that they give goes over the dom of g(x)...so i don't see how it would work

[Andiio]

Using the calc by defining the functions with the appropriate domain gives the graph with domain (- infinty, -5/2)

Yeah I tried that too...

This is really weird.

Applying the 'formula', dom fog should be dom g(x) which is (-∞,-5/2)

But my logic from working systematically is that it should R\{-5/2}, after checking my answer above and seeing that it was wrong

And the solutions say that it is (-∞,0)\{-5/2}

.....

[nacho]

Using the calc by defining the functions with the appropriate domain gives the graph with domain (- infinty, -5/2)

Yeah I tried that too...

This is really weird.

Applying the 'formula', dom fog should be dom g(x) which is (-∞,-5/2)

But my logic from working systematically is that it should R\{-5/2}, after checking my answer above and seeing that it was wrong

And the solutions say that it is (-∞,0)\{-5/2}

.....

its not R\{-5/2} because the domain of g is already restricted

[Andiio]

My reasoning is that:

x E (-∞,-5/2) which means g(x) E (-∞,0)

We need g(x) E R\{0},

Hence change it so that x E R\{-5/2} to fit all criteria.

If the final answer is indeed (-∞,-5/2), then what's wrong with my above reasoning?

Thanks guys!

[tony3272]

I've been thinking for ages that it's .

The range of g, is already a subset of the domain of f, and so you don't need to change it.

@Andiio: The range of the inner function only has to be a subset of the domain of the outer function, not equal to.

[brightsky]

My reasoning is that:

x E (-∞,-5/2) which means g(x) E (-∞,0)

We need g(x) E R\{0},

Hence change it so that x E R\{-5/2} to fit all criteria.

If the final answer is indeed (-∞,-5/2), then what's wrong with my above reasoning?

Thanks guys!

we don't need g(x) E R\{0}, rather g(x) must be a subset of dom f(x), and it already is.

[Andiio]

I guess I just got paranoid after initially getting (-∞,-5/2) and seeing it was 'wrong' in the solutions, so then did some unnecessary working :|



So in summary if fog is defined, then dom of fog = dom of g(x), for ALL CASES?


EDIT: whoops yeah that's what I meant. Thanks!

[brightsky]

not always. only when ran of g(x) is a subset of dom f(x). if you need to restrict the dom for fog to be defined, then the max dom would be something different, most likely a combination of the dom of g(x) and max dom of fog.