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