Question

Ok so I'm having trouble with these kinds of questions- I've always managed to make mistakes with these. they will be attached. any tips?

Answer 1

Sure. I can help.
g(f(x)) means that "g of f of x." So to find the solution, you look at g(x) and you substitute f(x) in wherever you see x.

Answer 2

I know that lol. For some reason though, when i substitute, i make some sort of error and my answers end up way off. Im sick and tired of it.

Answer 3

I can show you how to not make mistakes

Answer 4

So if I have
f(x) = 2x
g(x) = 1/x
Then f(g(x)) = 2(1/x) = 2/x
g(f(x)) = 1/(2x)

Answer 5

Give an example of one you messed up on maybe? There are kind of a lot of little things that you could be messing up.

Answer 6

Hey Smooth Math, that whiteboard feature is going to be cool

Answer 7

You're so right.

Answer 8

I just checked it out on the dev site

Answer 9

Yeah, I'm psyched about it. Should save a lot of effort and make using pictures a lot easier.

Answer 10

I'm quite surprised at how easy it was for them to incorporate it into this site. I think they did a good job

Answer 11

take the first problem on my attachement, \[x^2+5(\sqrt{x-3})\]

Answer 12

I dont remember where to start, i tend to over analyze these.. And somehow i end up doing something crazy...

Answer 13

x^2+5 is h(x) and \[\sqrt{x-3}\] is g(x)

Answer 14

And they want h(g(x)).
Be careful. h(g(x)) is not the same as h(x)*g(x)

Answer 15

You probably may need someone to explain it to you in more than just words though.

Answer 16

You have to plug g(x) in for x the same way you would plug in 3 for x if I asked you h(3)

Answer 17

OOOH i see what you are saying... Ive never liked these, in pre cal i always ended up making simple mistakes like that. which probably explains why my answers were always off. so how would that look like

Answer 18

That's a good explanation SmoothMath

Answer 19

Thanks, Hero. Do you get it, Liz?

Answer 20

so would it look like \[\sqrt{x-3} + 5 \] ? i hear what youre saying but Im afraid of making more mistakes.

Answer 21

Almost.
You start off with
h(x) = x^2 + 5
if I want h(3) I'll do:
(3)^2 + 5

Answer 22

Since I want h(g(x)) instead, I'll do
(g(x))^2 + 5
\[(\sqrt{x-3})^2 + 5\]

Answer 23

ok so you would have to square the \[\sqrt{x-3}\] which then bcomes just x-3 +5 ?

Answer 24

Yes!

Answer 25

so the answer would be x+2 ?

Answer 26

thanks though, you guys have helped me so much! :DDD

Answer 27

Yes, the answer is x + 2 good job

Answer 28

Yes. My pleasure. Just remember that you plug the whole equation in for x instead of multiplying the two together.
And then the other thing to remember is that if x gets multiplied by something or raised to a power, then the whole inside function gets operated on in that way.

Answer 29

YAY lol, ok so i think i can handle b and c on my attachment, but i have a question about D. I dont remember much on inverses. Care to explain a bit on that one?

Answer 30

Yes I do.

Answer 31

Haha ok.

Answer 32

Okay, so if I have f(x), it takes x values and outputs a single y value for each x.

Answer 33

The inverse equation of f(x) takes those y values as inputs, and it outputs the x that would give that y.

Answer 34

So if f-1(x) is the inverse function, and
f(1) = 2, then
f-1(2) = 1

Answer 35

Understand?

Answer 36

I see what you are saying, so for my problem, what work would i have to do for \[x^2 +5 \] ? would i have to just plug in numbers? Ugh i wish i remembered how to do these, i feel like such a bother.

Answer 37

It's actually pretty simple. I like to write the equation as
y = x^2+5
Then I switch x and y to get
x = y^2 + 5
Then I solve for y.
x-5 = y^2
y = sqrt(x-5)

This works because I want my outputs to be my inputs and vice versa.

Answer 38

Oh that really is simple, I think I'm just over thinking it and making it much harder then it seems. You have really cleared up a lot of my frustrations. It just looked intimidating at first.

Answer 39

And if you want to check your work, you can look and see what you get for f(f-1(x))
In this case I get
(f-1(x))^2 + 5 =
sqrt(x-5)^2 +5 =
x - 5 + 5 = x, which is what I should get.

Answer 40

Yes. Let me know if I can clear anything else up for you.

Answer 41

Thank you so much !!!!

Answer 42

for C i got 1/2 is that right. im currently trying to do B.

Answer 43

in b's case you would multiple f(x) by h(x) and then by -2?

Answer 44

Your answer for C isn't quite right. You got f(3) and what they want is f(f(3))

Answer 45

So you did the first step, which is to find f(3) and now you need to plug it in to f(x).
So f(f(3)) = f(1/2) = (1/2)/(1/2 + 3)

Answer 46

And for part B, that's just a fancy way of writing f(h(2))

Answer 47

oh ok so i just missed a step, atleast i was somewhat on the right track

Answer 48

f circle h means "f of h" and then what's in parenthesis afterwards is what goes inside h. It's not being multiplied.

Answer 49

Yes. You did the correct first step.

Answer 50

so in that case would you plug in \[(-2)^2 +5 \] to \[x/x+3 \]

Answer 51

Yes =)

Answer 52

Probably easiest to just solve that first thing first, then plug it in for x.

Answer 53

so the answer would be 3/4 if im not mistaken? or did i mess up the whole (-2)^2 vs. -2^2 ?

Answer 54

No you got it right =)

Answer 55

9/12 which reduces to 3/4

Answer 56

yay! thanks for being so patient, I know feel much more comfortable with these problems. I feel silly that i struggled with this last year when its really easy.

Answer 57

now*

Answer 58

It's not easy until you learn it =)

Answer 59

I'm glad I could help.

Answer 60

:)