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

Let $f: R - \{\frac{\alpha}{6}\} \rightarrow R$ be defined by $f(x) = \frac{5x + 3}{6x - \alpha}$. Then the value of $\alpha$ for which $(f \circ f)(x) = x$,for all $x \in R - \{\frac{\alpha}{6}\}$,is:

Suppose $f(x) = (x + 1)^2$ for $x \ge -1$. If $g(x)$ is the function whose graph is the reflection of the graph of $f(x)$ with respect to the line $y = x$,then $g(x)$ equals

Let $R$ denote the set of all real numbers. Let $f: R \rightarrow R$ and $g: R \rightarrow (0, 4)$ be functions defined by $f(x) = \log_e(x^2 + 2x + 4)$ and $g(x) = \frac{4}{1 + e^{-2x}}$. Define the composite function $h(x) = (f \circ g^{-1})(x)$,where $g^{-1}$ is the inverse of the function $g$. Then the value of the derivative of the composite function $h(x)$ at $x = 2$ is:

If $f(x) = \frac{x}{1 + x}$,then ${f^{-1}}(x)$ is equal to

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 $f$.)