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

Let $f:(0,1) \rightarrow R$ be defined by $f(x)=\frac{b-x}{1-b x},$ where $b$ is a constant such that $0 < b < 1$. Then

Let $f: R \rightarrow R$ be defined as $f(x)=10x+7$. Find the function $g: R \rightarrow R$ such that $g \circ f = f \circ g = I_{R}$.

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

If $f(x) = 3x - 5$,then ${f^{ - 1}}(x)$ is:

Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be differentiable functions such that $(f \circ g)(x) = x$. If $f(x) = 2x + \cos x + \sin^2 x$, then the value of $\sum_{n=1}^{99} g(1 + (2n - 1) \pi)$ is