# Thread: Solving functions with graph, inverses and etc.?

1. ## Solving functions with graph, inverses and etc.?

F(x) = { (3,5),(2,4),(1,7) }
H(x) = { (3,2),(4,3),(1,6) }
h(x) = sq. root (x-3)
k(x) = (x^2) + 5

Then I have a picture of the graph of f(x) and g(x) but not H & h

I'm not sure how to do these:

1/F(x), F^-1 (x), f^-1 (x)

2. Originally Posted by realintegerz
F(x) = { (3,5),(2,4),(1,7) }
H(x) = { (3,2),(4,3),(1,6) }
h(x) = sq. root (x-3)
k(x) = (x^2) + 5

Then I have a picture of the graph of f(x) and g(x) but not H & h

I'm not sure how to do these:

1/F(x), F^-1 (x), f^-1 (x)
Could you provide us with the complete unabrigded and unedited text of your problem? What you have posted doesn't make sense to me ...

3. For the others you can swap the x and y over then solve for y. Also remember that the range of the original function becomes the domain of its inverse

e.g.

$h(x)=\sqrt{x-3}$ , the range of this function is $[0,\infty)$

$y=\sqrt{x-3}$

$x=\sqrt{y-3}$

$y=x^2+3$

$\therefore h^{-1}(x)=x^2+3$ , the domain of this function is $[0,\infty)$

4. Originally Posted by earboth
Could you provide us with the complete unabrigded and unedited text of your problem? What you have posted doesn't make sense to me ...
Well it is how it is intended to be..

I got a graph with two functions on it, f and g, then I have those 4 equations for F(x), h(x), H(x), and k(x)

I'm just having trouble understanding what those 3 mean...because F(x) is a set of values and f(x) I don't have the equation for

5. Originally Posted by realintegerz
Well it is how it is intended to be..

I got a graph with two functions on it, f and g,

...
Is it possible for you to post the drawing? (As you may know it is more than 1000 words )

6. Oh I see what you mean... f and g are functions on a graph, I don't have the equations to them...

All I have given is what I have above.

7. That's what the problem looks like. Does it make a difference now?

I tried to explain the best I could but now you got a picture if it helps.

I still don't get how to solve these : 1/F(x), F^-1 (x), f^-1 (x)

Since F is coordinates and f is on the graph.

8. To list the points in 1/F(x), take the x-values from the points in F(x), and invert the corresponding y-values. For instance, the point (3, 5) in F(x) becomes the point (3, 1/5) in 1/F(x).

To list the points in the inverse of F(x), just apply the definition of the inverse and swap the x- and y-coordinates.

The graph shows that f(x) is a linear equation, so use what you've learned about slope and straight-line equations to find the equation for f(x). Then follow the process you've learned for finding the inverse of a function.

9. Ok so this is what I got:

1/F(x) = {(3, 1/5), (2, 1/4), (1, 1/7)}

F^-1(x) = {(5,3), (4, 2), (7, 1)}

f^-1(x) = 1 - x