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1 Answer
- az_lenderLv 72 months ago
(b) g(x) = pi*cos[(1/3)x] + pi.
h = (1/3), k = pi, m = pi.
(c) It may be easier to start (c) before (a).
y = f(x) = a*arcos[(x + b)/c] =>
y/a = arcos[(x + b)/c] =>
cos(y/a) = (x + b)/c =>
c*cos(y/a) - b = x.
Now it's easier (for me) to see the values of a,b,c by looking at the graph "sideways." However, it's hard to tell whether the "f" curve is supposed to represent a full half-cycle, as it doesn't really appear tangent to the y-axis in the problem. If it is intended to be tangent to the y-axis (i.e., vertical at y = 10pi), then a = 1/10, b = -15pi/2, and c = 15pi/2.
And the answer to (c) is
x = (15pi/2)*cos(y/10) + 15pi/2.
(a) x - 15pi/2 = (15pi/2)*cos(y/10) =>
y = 10*arcos[(x - 15pi/2)/(15pi/2)] = f(x) =>
a = 10, b = -15pi/2, c = 15pi/2.
