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31_Cal_Solution of Calculus_6e

# 31_Cal_Solution of Calculus_6e - f(g(x = f(cos x = cos x...

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f ( g ( x )) = f (cos x ) = cos x and g ( f ( x )) = g ( x ) = cos ( x ) . C01S04.017: If f ( x ) = sin x and g ( x ) = x 3 , then f ( g ( x )) = f ( x 3 ) = sin ( x 3 ) = sin x 3 and g ( f ( x )) = g (sin x ) = (sin x ) 3 = sin 3 x. We note in passing that sin x 3 and sin 3 x don’t mean the same thing! C01S04.018: If f ( x ) = sin x and g ( x ) = cos x , then f ( g ( x )) = f (cos x ) = sin(cos x ) and g ( f ( x )) = g (sin x ) = cos(sin x ). C01S04.019: If f ( x ) = 1 + x 2 and g ( x ) = tan x , then f ( g ( x )) = f (tan x ) = 1 + (tan x ) 2 = 1 + tan 2 x and g ( f ( x )) = g (1 + x 2 ) = tan(1 + x 2 ). C01S04.020: If f ( x ) = 1 x 2 and g ( x ) = sin x , then f ( g ( x )) = f (sin x ) = 1 (sin x ) 2 = 1 sin 2 x = cos 2 x and g ( f ( x )) = g (1 x 2 ) = sin(1 x 2 ) . Note: The answers to Problems 21 through 30 are not unique. We have generally chosen the simplest and most natural answer. C01S04.021: h ( x ) = (2 + 3 x ) 2 = ( g ( x )) k = f ( g ( x )) where f ( x ) = x k , k = 2, and g ( x ) = 2 + 3 x . C01S04.022: h ( x ) = (4 x ) 3 = ( g ( x )) 3 = f ( g ( x )) where f ( x ) = x k , k = 3, and g ( x ) = 4 x . C01S04.023: h ( x ) = (2 x x 2 ) 1 / 2 = ( g ( x )) 1 / 2 = f ( g ( x )) where f ( x ) = x k , k = 1 2 , and g ( x ) = 2 x x 2 . C01S04.024: h ( x ) = (1 + x 4 ) 17 = ( g ( x )) 17 = f ( g ( x )) where f ( x ) = x k , k = 17, and g ( x ) = 1 + x 4 . C01S04.025: h ( x ) = (5 x 2 ) 3 / 2 = ( g ( x )) 3 / 2 = f ( g ( x )) where f ( x ) = x k , k = 3 2 , and g ( x ) = 5 x 2 . C01S04.026: h ( x ) = (4 x 6) 1 / 3 4 = (4 x 6) 4 / 3 = ( g ( x )) 4 / 3 = f ( g ( x )) where f ( x ) = x k , k = 4 3 , and g ( x ) = 4 x 6. Alternatively, h ( x ) = ( g ( x )) 4 = f
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