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Mathematics 18 Online
OpenStudy (anonymous):

Find h(f(x)), in terms of x for the following functions, given that: f(x)= x^2 + x h(x)= x/1-x

OpenStudy (anonymous):

Alright, let's see if I remember how to do these... ready to work on this together?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

h(f(x)) means plug the function f(x) into the function h(x) as the input. So, h(f(x)) = h(x^2+x)...

OpenStudy (anonymous):

Replace x in the h(x) function with x^2+x :)

OpenStudy (anonymous):

Alright, first thing we want to do is understand what h(f(x)) means.

OpenStudy (anonymous):

What this means is "do f(x) first, and then plug that into h(x)"

OpenStudy (anonymous):

Does that make sense?

OpenStudy (anonymous):

yeah so i get x^2+x/1-(x^2+x)

OpenStudy (anonymous):

Perfect.

OpenStudy (anonymous):

but then how would i simplify it?

OpenStudy (anonymous):

I'll have to grab some pen and paper. I'll be right back.

OpenStudy (anonymous):

you don't really need to much. best you can get is\[(x^{2}+x)/(1-x-x^{2})\]after distributiong the negative on the bottom

OpenStudy (anonymous):

the denominator doesn't factor, so you can't go any further.

OpenStudy (anonymous):

also, did my answer to your other problem help, jerni?

OpenStudy (anonymous):

Yep, mtbender's correct. It *can* be modified a bit, but this is as condensed as we can make it.

OpenStudy (anonymous):

personally, i'd leave it without distributing the negative. needless "simplification" when it doesn't get you anything is just a waste of time and opens the door for errors. :)

OpenStudy (anonymous):

yes it did thank you:)

OpenStudy (anonymous):

my pleasure. :)

OpenStudy (anonymous):

ah i get it now thank you guys so much!

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