The point of intersection can therefore be found by solving the system of equations □ = □ + 3 6 1 2 5, □ = □. Since point □ is a point of intersection of □ = □ ( □ ) and □ = □ ( □ ) , it must lie on the line □ = □. Recall that the graph of □ = □ ( □ ) is mapped onto □ = □ ( □ ) by a reflection in the line □ = □. Hence, the □-coordinate of the point □ is 3 6 1 2 5. The □-intercept of □ ( □ ) is therefore given by 3 6 1 2 5, 0 This function fails the horizontal line test as shown below, so it is not one-to-one. Further, if a function has no inverse, it might be possible to restrict the domain of that function so that this new function does have an inverse.įor instance, consider the function □ ( □ ) = □ whose graph is shown. Similarly, since the inputs of □ are the outputs of □ , the domain of □ is the range of □ . Since the outputs of □ are the inputs of □ , the range of □ is also the domain of □ . In fact, we might even deduce more information about the domain and range of functions and their inverses. Similarly, if the coordinate of the □-intercept on the graph of □ = □ ( □ ) is ( □, 0 ) for some real □, the image of this point on the graph of □ = □ ( □ ) is ( 0, □ ). It follows that the converse must also be true. In our previous example, we saw that if the coordinate of the □-intercept on the graph of □ = □ ( □ ) is ( 0, □ ) for some real □, the image of this point on the graph of □ = □ ( □ ) is ( □, 0 ).
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