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f(x)=23x+27 
g(x)=12x-d 

If (f^-1(g^-1(x)))= (g^-1(f^-1(x))), find d. 

How do I do this please?


Sagot :

[tex]f(x)=23x+27\\ y=23x+27\\ 23x=y-27\\ x=\frac{1}{23}y-\frac{27}{23}\\ f^{-1}(x)=\frac{1}{23}x-\frac{27}{23}\\\\ g(x)=12x-d\\ y=12x-d\\ 12x=y+d\\ x=\frac{1}{12}y+\frac{1}{12}d\\ g^{-1}(x)=\frac{1}{12}x+\frac{1}{12}d\\\\ [/tex]

[tex]\frac{1}{23}(\frac{1}{12}x+\frac{1}{12}d)-\frac{27}{23}=\frac{1}{12}(\frac{1}{23}x-\frac{27}{23})+\frac{1}{12}d\\ \frac{1}{276}x+\frac{1}{276}d-\frac{27}{23}=\frac{1}{276}x-\frac{27}{276}+\frac{1}{12}d\\ \frac{1}{276}d-\frac{1}{12}d=-\frac{27}{276}+\frac{27}{23}=\\ \frac{1}{276}d-\frac{23}{276}d=-\frac{27}{276}+\frac{324}{276}=\\ -\frac{22}{276}d=\frac{297}{276}=\\ -22d=297\\ d=-\frac{297}{22} [/tex]