solve for a b c and d exponential functions

Question

suppose 𝑓(𝑥)=𝑥+4and𝑔(𝑥)=2𝑥−5 then

A.) (𝑓∘𝑔)(𝑥)=

B.) (𝑓∘𝑔)^−1(𝑥)=

C.) (𝑔^−1∘𝑓^−1)(𝑥)=

D.) (𝑔^-1∘f^-1)(x)=

Answer 1

(𝑓∘𝑔)(𝑥) means that you will replace any variables in the function f with the function of g. 𝑓^−1 and 𝑔^−1 are the reciprocal functions, meaning that you will take 1/f or 1/g.

A.) (𝑓∘𝑔)(𝑥)

=(f g(x))

=(2x-5)+4

=2x-1

B.) (𝑓∘𝑔)^−1(𝑥)

=1/(𝑓∘𝑔)(𝑥)

=1/(2x-1)



C.) (𝑔^−1∘𝑓^−1)(𝑥)=

=1/g(1/f(x))

=1/(2(1/(x+4))-5)

=1/((2/(x+4))-((5x-20)/(x+4)))

=1/((2-5x-20)/(x+4))

=1/((18-5x)/(x+4))

=(x+4)/(-18-5x)

C and D are the same, so I am assuming one of them should have been (f^-1 ∘ g^-1)(x)

D.) (f^−1∘g^−1)(𝑥)

=1/(2x-5) + 4

=1/(2x-5) + (8x-20)/(2x-5)

=(1+8x-20)/(2x-5)

=(8x-19)/(2x-5)