If $f(x) = x^2 - x + 5, x > \frac{1}{2},$ and $g(x)$ is its inverse function,then $g'(7)$ equals

If $g$ is the inverse of $f$ and $f^{\prime}(x)=\frac{1}{1+x^{2}}$,then $g^{\prime}(x)$ is equal to

$f(x) = \sin x + \cos x, g(x) = x^2 - 1$. Then $g(f(x))$ is invertible if:

The inverse of $y = 5^{\log x}$ is

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 $f$.)

Let $f : (4, 6) \to (6, 8)$ be a function defined by $f(x) = x + [\frac{x}{2}]$ (where $[.]$ denotes the greatest integer function),then $f^{-1}(x)$ is equal to