If $f: \mathbb{R} \to \mathbb{R}$ is defined by $f(x) = 3x - 4$,then ${f^{ - 1}}: \mathbb{R} \to \mathbb{R}$ is

Consider $f: R_{+} \rightarrow [-5, \infty)$ given by $f(x) = 9x^{2} + 6x - 5$. Show that $f$ is invertible with $f^{-1}(y) = \frac{\sqrt{y+6}-1}{3}$.

Let $f: X \rightarrow Y$ be an invertible function. Show that $f$ has a unique inverse.
(Hint: Suppose $g_{1}$ and $g_{2}$ are two inverses of $f$. Then for all $y \in Y$,$f \circ g_{1}(y) = I_{Y}(y) = f \circ g_{2}(y)$. Use the one-one property of $f$.)

If the functions $f$ and $g$ are defined by $f(x) = 3x - 4$ and $g(x) = 2 + 3x$ for $x \in R$, then $g^{-1}(f^{-1}(5))$ is equal to

Consider $f: \{1, 2, 3\} \rightarrow \{a, b, c\}$ given by $f(1) = a, f(2) = b$ and $f(3) = c$. Find $f^{-1}$ and show that $(f^{-1})^{-1} = f$.

If $f: R \rightarrow R$ is a mapping defined by $f(x)=x^{3}+5$,then $f^{-1}(x)$ is equal to