By Munkres J.R.

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Since the graph is symmetric with respect to the jt-axis, (x, -y) is also on the graph, and, therefore, since the graph is symmetric with respect to the y-axis, (—x, —y) is also on the graph. Thus, the graph is symmetric with respect to the origin. 50 is false. The graph of the odd function fix) = x is symmetric with respect to the origin. However, (1,1) is on the graph but (—1,1) is not; therefore, the graph is not symmetric with respect to the y-axis. It is also not symmetric with respect to the x-axis.

Thus, the domain is [-5,1]. For* in the domain, 9 > 9 - (x + 2)2 > 0, and, therefore, the range will be [0,3]. 90 Show that the product of two even functions and the product of two odd functions are even functions. If / and g are even, then f(~x)-g(-x) = f(x)-g(x). 91 Show that the product of an even function and an odd function is an odd function. Let /be even and g odd. 92 Then f(-x)-g(-x) =/(*)• [-g«] = or —x — g(—y). Thus, g(—y) = [/(-*) =/(*)]• Find an equation of the new curve C* when the graph of the curve C with the equation reflected in the x-axis.

Fig. 7-1 x = 0 is a point of discontinuity because lim f(x) does not exist, x = 1 is a point of discontinuity because lim f ( x ) * f ( l ) [since lim/(jt) = 0 and /(I) = 2]. 3 Determine the points of discontinuity (if any) of the function f(x) such that f(x) = x2 if x =£ 0 and f(x) - x if x>0. f(x) is continuous everywhere. In particular, f(x) is continuous at x = 0 because /(O) = (O)2 = 0 and lim f(x) = 0. 4 Determine the points of discontinuity (if any) of the function/(*) such that f(x) = 1 if x^O and /(jt)=-l if x<0.