finding f(x) given g(x) and (fog)(x) and vice versa

[victini]

I've been stuck on these homework problems for hours. I'd appreciate some guidance.

1.) Find f(x), given g(x) and (fog)(x):

g(x)= 1/x
(fog)(x)=x

I've tried a lot of wrong things.

f(1/x)=x

2.) Find g(x), given f(x) and (fog)(x):

f(x)= 1+(1/x)
(fog)(x)= x

I've tried to isolate g(x) with 1+ 1/g(x) = x but I'm not coming up with anything correct. g(x) always ends up attached to something.

1+ 1/g(x) =x
1+1 = x(g(x))
2=x(g(x))

1+ 1/g(x) =x
1/g(x) = x-1

 

[stapel]

I've been stuck on these homework problems for hours. I'd appreciate some guidance.

I'm not aware of any "formula" for this sort of exercise. You just kind of have to use your experience from your coursework, and some fiddling around, to find something that works.

1.) Find f(x), given g(x) and (fog)(x):

g(x)= 1/x
(fog)(x)=x

You've got a function that inverts, and you've got a composition that takes you back to just the original variable. Back in algebra (you'd originally posted this to "Calculus"), you learned about composition and inverses; specifically, you learned that inverse functions, when composed, take you back to "just x". So what then must f(x) be, if f composed with g ends up with just x?

I've tried a lot of wrong things.

Such as...?

2.) Find g(x), given f(x) and (fog)(x):

f(x)= 1+(1/x)
(fog)(x)= x

Do the regular process (here) for finding the inverse, since obviously this is what you need in order to get the given result.

 

[victini]

You've got a function that inverts, and you've got a composition that takes you back to just the original variable. Back in algebra (you'd originally posted this to "Calculus"), you learned about composition and inverses; specifically, you learned that inverse functions, when composed, take you back to "just x". So what then must f(x) be, if f composed with g ends up with just x?

This is for a calculus course, sorry for putting it in the wrong spot.

I know the inverse is multiplication. Idk what to do next. If I take 1/x and multiply by x its I'll end up with x. the problem is f(x) can't be x(x), because that would be (1/x)(1/x) which doesn't give me x.

Do the regular process (here) for finding the inverse, since obviously this is what you need in order to get the given result.

Okay I tried that, maybe I'm not understanding?

y = 1+ 1/x
xy = 1+1

x = 2/y
y = 2/x

When I test that at f(g(2))

1+ 1/ (2/2)

1+ 1/1

1+1 =2


When I test that f(g(4)

1+ 1/ (2/4)

1+ 1/ .5)

1 + 2 = 3

3 =/= 4 it did not work

 

[stapel]

I know the inverse is multiplication.

I'm sorry, but I'm not sure what you mean by this...?

Try doing the suggested thing: You have a function and you have a result of composition. The result is "just x", which means the composition must have been with the inverse function. So find that inverse function, using the process explained and illustrated in the lesson at the link.

You are given that g(x) = 1/x. Follow the standard procedure:

. . . . .i) Replace the function name with "y".
. . . . .ii) Solve the resulting equation for "x=".
. . . . .iii) Swap "x" and "y".
. . . . .iv) Rename "y" as the inverse function.

Example:

. . . . .g(x) = 2x^(1/3) = (2) * (cube root of x)
. . . . .(fog)(x) = x

. . . . .then f(x) must be inverse of g(x), so find inverse of g(x):

. . . . .g(x) = 2x^(1/3)
. . . . .i) y = 2x^(1/3)
. . . . .ii) y/2 = x^(1/3)
. . . . .(y^3)/8 = x
. . . . .iii) (x^3)/8 = y
. . . . .iv) g^-1(x) = (x^3)/8

since f(x) is inverse function, then f(x) = g^-1(x) = (x^3)/8

Follow this procedure.

y = 1+ 1/x
xy = 1+1

How did you get this? Please show all of your steps, including the step where you multiplied through (I think?) by x:

. . . . .\(\displaystyle y\, =\, 1\, +\, \dfrac{1}{x}\)

. . . . .\(\displaystyle (y)x\, =\, \left(1\, +\, \dfrac{1}{x}\right)x\)

. . . . .\(\displaystyle xy\, =\, \left(1\right)x\, +\, \left(\dfrac{1}{x}\right)x\)

What did you do next? Thank you! :wink:

 

[victini]

Thanks for your patience. I think I got the answers last night, they are 1/x and 1/x-1. They popped into my head and I thought "let's try it". Luck? I feel like I didn't learn anything
The solve for x, I think, is messing me up. Algebra skills are rusty. But I see how you got ..xy=(1)x+(1x)x

This is the closest I've gotten:

xy = 1x + (1/x)x Can I distribute the x?
xy = x + x/2x
xy(2x) = x + x
xy(2x) = 2x divide by 2x?
xy = 1
x = 1/y

Swap x and y

y = 1/x ?
​







previous tries (to show effort):

xy = 1x + (1/x)x
xy = x + x
y = 2x/x
Start over

xy = 1x + (1/x)x
xy - x = (1/x)x
xy - x = x
x - x = x/y
0 = x/y
y(0) = x
obviously wrong

xy = 1x + (1/x)x
xy - x =(1/x)
I don't know.

 

[stapel]

I feel like I didn't learn anything
The solve for x, I think, is messing me up. Algebra skills are rusty.

That's why you were given a link to a lesson: so you could brush up on your skills, and be sure that you learned how to do this. You might want to go now and study that lesson; you will need to know how to do this stuff.

But I see how you got ..xy=(1)x+(1x)x

This is the closest I've gotten:

xy = 1x + (1/x)x Can I distribute the x?

Yes. This was something they were supposed to have taught you in your previous algebra course. To learn how to work with parentheses, try here. To learn the basic steps for solving one-variable equations (which one uses at this stage of the game), try here, here, and here. To learn some of the basics of solving equations with more than one variable, try here.

xy = x + x/2x

How did "(1/x)(x/1)" turn into "2"?

xy(2x) = x + x

How did you multiply the first term on the right-hand side, being "x", by 2x and end up with "x"? That is, how did "(x)(2x)" become "x"?

Instead, try following the earlier instructions by showing all of your steps. By not skipping steps, you may be able more clearly to "see" what is going on.