Find the reflection of the line y = mx + b in the origin.
e replace x by -x and y by -y, obtaining -y = -mx+b, that is, y = mx-b. Thus, the y-intercept
changes to its negative and the slope remains unchanged.
Show geometrically that when the graph of a one-one function is reflected in the 45° line y = x, the result is the
graph of the inverse function.
It is evident from Fig. 5-26 that right triangles ORP and OR'P' are congruent. Hence,
Thus the locus of P' is the graph of x as a function of y; i.e., of x = f~i(y). Note that because y = f(x)
meets the horizontal-line test (/being one-one), x = f~(y) meets the vertical-line test.
Fig. 5-26
5.101 Graph the function f(x) = V|*-l|-l.
The complement of the domain is given by |jt-l| the domain consists of all x such that x < 0 or x a 2. Case 1. x a 2. Then y = Vx^2, y
2 = * - 2.
So, we have the top half of a parabola with its vertex at (2,0) and the jt-axis as axis of symmetry. Case 2.
x < 0. Then, y = V^x, y
2 = -x. So, we have the top half of a parabola with vertex at the origin and with
as axis of symmetry. The graph is shown in Fig. 5-27.
Fig. 5-27
34
5.99
5.100
CHAPTER 5
andd d
R'P'=RP=y
OR'=OR=x