If f¹ is supposed to say f⁻¹ : rewrite f(4x + 5) as an "obvious" function of 4x + 5, then replace 4x + 5 with x. By "obvious", I mean make it clear how 4x + 5 is the argument to f.
[tex]f(4x+5) = 12x+18 = 3 (4x + 6) = 3 (4x + 5 + 1) = 3 (4x + 5) + 3[/tex]
Then swapping out 4x + 5 for x gives
[tex]f(x) = 3x + 3[/tex]
The inverse of f(x), if it exists, is a function f⁻¹(x) such that
[tex]f\left(f^{-1}(x)\right) = x[/tex]
Evaluate f at f⁻¹ and solve for f⁻¹ :
[tex]f\left(f^{-1}(x)\right) = 3 f^{-1}(x) + 3 = x \implies \boxed{f^{-1}(x) = \dfrac{x-3}3}[/tex]
If f¹ instead means f' (as in the first derivative of f) : earlier we found f(x) = 3x + 3, and differentiating this is trivial.
[tex]f(x) = 3x+3 \implies f'(x) = 3[/tex]
