I recommend that you sketch a graph of this parabola and include it in your solution to help visualize the problem. Just make a quick table of values. As I suggested in my previous email about sign diagrams, don't forget to choose negative x values as well as
positive x values when making a table. A good default set of x values is 0, 1, -1, 2, and -2 when you want to make a quick graph.
Part (a)
The vertical line test checks if a graph is a function or not. If a vertical line can never pass through more than one point on the graph, the graph represents a function
The horizontal line test checks if a
graph's inverse would be a function. If a horizontal line can never pass through more than one point on the graph, then its inverse would be a function.
A graph is a one-to-one function if and only if it passes both the vertical and horizontal line tests. That proves that for any one value of x there is only one value of y on the graph. The function is
one-to-one.
Sketch the graph of the parabola, and it is obvious that it fails the horizontal line test. So you must restrict the domain to make it one-to-one. The most logical thing to do is to restrict the domain to all the x values from the vertex and larger.
Part (b)
I show you how to find the inverse of a function in Lesson 8 of my book (just
after question 4 in the lecture, page 230). I like to immediately have x and y change places, then proceed to isolate y.
Once you have isolated y, make sure you replace y with f^-1 (x), the f-inverse symbol.
Since an inverse changes the x and y around, it is sort of rotating a graph sideways (and flipping it). What was vertical becomes horizontal, and vice-versa.
Thus, a graph that passes the horizontal line test, will have an inverse that passes the vertical line test, proving that the inverse will be a function. Only one-to-one functions will have inverses that are also function.
Part (d)
The domain and range for f(x) should be obvious from the sketch you made earlier in the question (but, remember, you must use the restricted domain you identified in part (a) for the
one-to-one function). Then, the domain and range of the inverse function are easy. The domain of f is the range of f-inverse, and the range of f is the domain of f-inverse.