Question #86e50

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Answer 1 · 2017-03-28T09:48:08

#f(x) = | sqrt (x + 9) |#

#f[g(x)] = | 2 x - 3 |#, #g(x) = 4 x^2 -12 x#

#f[4 x^2 -12 x] = | 2 x - 3 |# #->a#

Assuming
#y = 4 x^2 -12 x# #->f[4 x^2 -12 x] = f(y)#
#y = 4 (x^2 -3 x)#
#y/4 = (x^2 -3 x)#
solve using comleting a square
#y/4 = (x -3/2)^2 -(3/2)^2#

#y/4 + 9/4= (x -3/2)^2 #

#+- sqrt (y/4 + 9/4)= x -3/2#

# 3/2 +-sqrt (y + 9)/2 = x #

# (3 +-sqrt (y + 9))/2 = x #

plug #y# and #x = (3 +-sqrt (y + 9))/2 # in se #a#

#f(y) = | cancel2 ((3 +-sqrt (y + 9))/cancel2) - 3|#

#f(y) = | 3 +-sqrt (y + 9) - 3|#

#f(y) = | +-sqrt (y + 9) |#

since it result always +ve value (absoluted). Therefore,
#f(y) = | sqrt (y + 9) |#
#f(x) = | sqrt (x + 9) |#