Solve gf(x) = 1?

Question

#f(x) = (2x^2+3x-14)/(x^2-5x+6)#

#g(x)= (12-x^2)/x#

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Answer 1 · 2018-05-23T23:35:15

#g@f(16) =1# and #g@f(2) =1#

First we need to find the value where g(x) = 1

#g(x)= (12-x^2)/x#

# 1=(12-x^2)/x#

# x=12-x^2#

#x^2 + x -12=0#

#(x+4)(x-3)=0#

#x={3, -4}#

now we need to solve #f(x) = 3# or #f(x) = -4#, let's try 3:

#f(x) = (2x^2+3x-14)/(x^2-5x+6)#

#3 = (2x^2+3x-14)/(x^2-5x+6)#

#3(x^2-5x+6) = 2x^2+3x-14#

#3x^2-15x+18 = 2x^2+3x-14#

#3x^2-15x+18 -2x^2-3x+14=0#

#x^2-18x+32=0#

#(x - 2) (x - 16)=0#

#x={2, 16}#

There may be 2 other solutions using:

#-4 = (2x^2+3x-14)/(x^2-5x+6)#