Answer
(a) $f^{-1}(x) = \frac{1}{4}x + \frac{3}{4}$
(b) Both $f(f^{-1}(x))$ and $f^{1}(f(x))$ result in $x$.
Work Step by Step
(a) $$f(x) = 4x - 3$$ $$y = 4x - 3$$ $$x = 4y - 3$$ $$x + 3 = 4y$$ $$\frac{1}{4}x + \frac{3}{4} = y$$ $$\frac{1}{4}x + \frac{3}{4} = f^{-1}(x)$$
(b)
$f(f^{-1}(x))$ $$4(\frac{1}{4}x + \frac{3}{4}) - 3$$ $$x + 3 - 3$$ $$x$$
$f^{1}(f(x))$ $$\frac{1}{4}(4x - 3) + \frac{3}{4}$$ $$x - \frac{3}{4} + \frac{3}{4}$$ $$x$$
