Understanding inverse functions

[admiralgman]

I need some help understanding inverse functions, we've had a 4-page chapter covering the basics of inverse functions and I understand that.

But now we have suddenly gotten these task that I don't understand how to solve, I've read the part on inverses several times, but I still don't understand.

To explain my trouble, here are two examples from my textbook that covers what I am struggeling with:

1:Calculate the inverse of the given functions in terms of f^-1

g(x)=f(x)-2

(answer to this should be g^-1(x)=f^-1(x+2))

2: Find g^-1(1) if g(x)=x³+x-9

(answer to this should be 2)


As you see I know the answers, this because they are in the back of my textbook. But I want some help understanding how you get from that point A to B
Also, it would be nice if you could show it in a step-by-step way as I am (already) having some trouble understanding the technical terms in english, considering I am from Norway

As a last bit of information; I will be gone for two hours or so now(Im a busy man), but it would be nice to come back here and see something that would help me understand "advanced" inverses.

 

[Mentallic]

Welcome to PF!

1:Calculate the inverse of the given functions in terms of f^-1

g(x)=f(x)-2

(answer to this should be g^-1(x)=f^-1(x+2))

For this one, it would probably help if you drew some random function f(x) and to keep things simple, have it be entirely below the line y=x. Draw a dotted line y=x for reference, and then draw the inverse function g(x) = f^-1(x) which is a reflection of f(x) in the line y=x.

Now, if you draw f(x)-2, that's simply a shift of f(x) down 2 units, and so if you reflect f(x)-2 across the y=x line (which is also the same as considering what the inverse function is) then how is this new function shifted in relation to g(x)? Is it shifted down, up, left, right?

2: Find g^-1(1) if g(x)=x³+x-9

(answer to this should be 2)

If you were supposed to find g(2), then what do you get?

 

[D H]

2: Find g^-1(1) if g(x)=x³+x-9

(answer to this should be 2)

I'm only going to help with this second example as Mentallic already gave a nice way to look at the first example.

Rhetorical question: What does g^-1(y) mean? The y is some value in the range (image) of g(x), and g^-1(y) maps from this range back to the domain of g(x) in such a way that g^-1(g(x))=x and g(g^-1(y))=y.

What you are looking for here is some x such that g(x)=1. In other words, you need to solve x³+x-9=1, or x³+x-10=0. Solving this isn't that hard. Just use the rational root theorem. The only possible rational roots are ±10, ±5, ±2, and ±1. You should be able to eliminate all but ±2 by inspection. x-2 divides x³+x-10, so the solution is x=2.

 

[HallsofIvy]

I need some help understanding inverse functions, we've had a 4-page chapter covering the basics of inverse functions and I understand that.

But now we have suddenly gotten these task that I don't understand how to solve, I've read the part on inverses several times, but I still don't understand.

To explain my trouble, here are two examples from my textbook that covers what I am struggeling with:

1:Calculate the inverse of the given functions in terms of f^-1

g(x)=f(x)-2

By the definition of "inverse function" if y= f(x) then x= f^-1(y). Similarly, if y= g(x) then x= g^-1(x).

Let y= g(x)= f(x)- 2. Then f(x)= y+ 2 so that x= f^-1(y+ 2). Writing that in terms of the variable x, g^-1(x)= f^-1(x+ 2)

(answer to this should be g^-1(x)=f^-1(x+2))

2: Find g^-1(1) if g(x)=x³+x-9

If g(x)= x³ + x- 9, then g(2)= 2³+ 2- 9= 8+ 2- 9= 1
so that g^-1(1)= 2.

(answer to this should be 2)


As you see I know the answers, this because they are in the back of my textbook. But I want some help understanding how you get from that point A to B
Also, it would be nice if you could show it in a step-by-step way as I am (already) having some trouble understanding the technical terms in english, considering I am from Norway

As a last bit of information; I will be gone for two hours or so now(Im a busy man), but it would be nice to come back here and see something that would help me understand "advanced" inverses.