1.) f(x)=x-7/x+3 and g(x) = -3x-7/x-1
2.)Verify the identity. cos 4x + cos 2x = 2 – 2 sin^2 2x – 2 sin^2 x
2 Answers
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1)
You need to put parenthesis in your question. Did you mean g(x) = (-3x-7)/(x-1)?
f(x) = (x-7) /(x+3)
f(g(x)) = f [ (-3x-7) /(x-1)]
Replace x with (-3x-7)/(x-1) in f
Evaluate (-3x-7)/(x-1) by replacing x with (x-7)/(x+3)
evaluate the numerator (-3x-7):
– 3 (x-7)/(x+3) -7 = (-3x+21)/(x+3) – 7
= [(-3x+21) -(7x+21)] /(x+3) = (-3x+21-7x-21)/(x+3) = -10x/(x+3) ——(1)
evaluate the denominator (x+3)
(-3x-7) /(x+3) +3
= [ (-3x-7) + 3(x-1)] /(x+3) = (-3x-7+3x-3)/(x+3) =-10/(x+3) ——–(2)
(1)/(2) = -10x/-10 = x
We have shown that f(g(x)) = x
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g(x) = [(-3x-7) /(x-1)]
numerator of g(f(x)) = [ -3( (x-7)/(x+3) – 7] = [-3x+21-7(x+3)]/(x-1) = (-3x+21-7x-21)/(x-1) = -10x/(x-1)
denominator of g(f(x)) = (x-7)/(x+3)-1 = ((x-7)-x-3)/(x-1) = -10/(x-1)
divide = -10x/(x-1)/-10/(x-1) = x
f(g(x)) = g(f(x))
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2)
cos 4x +cos 2x = 2 cos ^2 2x -1 + 2 cos^2 x -1
= 2(1 – sin^2 2x ) -1 +cos 2x
=1-2sin^2 2x + 1-2sin^2 x
=2 – 2sin^2 2x – 2 sin^2 x
=
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Need ()
assuming I use the right ()
f(x)= (x-7)/(x+3)
g(x) = (-3x-7)/(x-1)
f(g(x)) = ((-3x-7)/(x-1)-7)/((-3x-7)/(x-1)+3 )
distribute then simplify
= x
g(f(x)) = (-3(x-7)/(x+3)-7)/((x-7)/(x+3)-1)
=x
