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Mathematics 18 Online
OpenStudy (anonymous):

What is the value of f[g(5)] for the functions f(x) = 2x + 1 and g(x) = 2x - 5? (Type only the value of your final answer, no variables.)

OpenStudy (shadowfiend):

f[g(5)] is a 2-step process. You work from the inside in. So first off you have to find g(5). Can you do that?

OpenStudy (anonymous):

not really :( if you can teach this stuff to me thatd be great

OpenStudy (shadowfiend):

Absolutely!

OpenStudy (shadowfiend):

So when I say g(5), we're referring to g as a “function”. Above, the function is defined as: g(x) = 2x - 5 That means that, given a number x, to get the value of g(x), you multiply that number by 2 and then subtract 5.

OpenStudy (anonymous):

IS it always by 5 that you subtract?

OpenStudy (shadowfiend):

No, that's just how g(x) is defined in this particular case.

OpenStudy (anonymous):

So it would be 2x5-5

OpenStudy (shadowfiend):

So for g(x) the way it's pointed out above, we would have: \[\begin{align} g(1) &= 2\cdot 1- 5 &= 2 - 5 &= -3\\ g(2) &= 2\cdot 2- 5 &= 4 - 5 &= -1\\ g(3) &= 2\cdot 3- 5 &= 6 - 5 &= 1\\ g(4) &= 2\cdot 4- 5 &= 8 - 5 &= 3 \end{align}\]

OpenStudy (shadowfiend):

Yep, g(5) would be exactly that! Which gives us?

OpenStudy (anonymous):

5 again

OpenStudy (shadowfiend):

Exactly :)

OpenStudy (shadowfiend):

So now we know that g(5) = 5. But we're looking for f[g(5)]. Once we know g(5) = 5, we can put it in wherever g(5) shows up. In this case: f[g(5)] = f(5)

OpenStudy (shadowfiend):

Now we can plug 5 into the definition of f(x). Can you do that?

OpenStudy (anonymous):

f(5) like you already did wouldnt it be?

OpenStudy (shadowfiend):

Not quite, f(x) is 2x + 1. So we plug 5 into that equation now.

OpenStudy (anonymous):

2*5+1

OpenStudy (anonymous):

=11

OpenStudy (shadowfiend):

Exactly! :)

OpenStudy (shadowfiend):

So f[g(5)] = f(5) = 11

OpenStudy (anonymous):

now what xD

OpenStudy (anonymous):

hello ?

OpenStudy (shadowfiend):

Sorry! Haha. That's it? That's the answer to the question :)

OpenStudy (anonymous):

o ok thank you lol thought there was more to it

OpenStudy (shadowfiend):

Nope :) Not that I can see, anyway.

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