SOLUTION: A2. Suppose the point (1, 2) lies on the graph of the function f. (a) Find a point on the graph of the function {{{g(f(x))}}}, where {{{g(y)=y-1+arctan (y/2)}}}. (b) Find two poi

Algebra ->  Testmodule -> SOLUTION: A2. Suppose the point (1, 2) lies on the graph of the function f. (a) Find a point on the graph of the function {{{g(f(x))}}}, where {{{g(y)=y-1+arctan (y/2)}}}. (b) Find two poi      Log On


   



Question 1035254: A2. Suppose the point (1, 2) lies on the graph of the function f.
(a) Find a point on the graph of the function g%28f%28x%29%29, where g%28y%29=y-1%2Barctan+%28y%2F2%29.
(b) Find two points on the graph of the function f%28h%28x%29%29, where h(x) = x^2 −πx+ (cos^2)x.
Thankyou for looking over. Please provide some steps to steps instructions if possible.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose the point (1, 2) lies on the graph of the function f.
(a) Find a point on the graph of the function g(f(x)), where g(y)=y-1+arctan (y/2).
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g[f(1)] = g(2) = 2-1+arctan(2/2) = 2-1+pi/4 = 1+pi/4
Point:: (1,g(f(1)) = (1,1+pi/4)
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(b) Find two points on the graph of the function f(h(x)),
where h(x) = x^2 −πx+ (cos^2)x
f[h(x)] = f[x^2-pi*x+cos^2(x)]
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Remember f(1) = 2, so::
If x^2-pi*x+cos^2(x) = 1
then x^2-pi = 1-cos^2(x)
and x^2-pi*x = sin^2(x)
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So f[x^2-pi*x+cos^2(x)] = f[sin^2+cos^2] = f(1) = 2
Point:: (x,f[h(x)]) = (1,2)
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If x = pi
then x^2-pi*x+cos^2(x) = pi^2 - pi^2 + 1 = 1
Point:: (x,f[h(x)] (pi,2)
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Cheers,
Stan H.
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Thankyou for looking over. Please provide some steps to steps instructions if possible.