$f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions such that $f(x)=2x-3$ and $g(x)=x^3+5$. Then,$(f \circ g)^{-1}(-9)$ is

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=x-[x]$ and $g(x)=[x]$ for $x \in R$,where $[x]$ is the greatest integer not exceeding $x$,then for every $x \in R, f(g(x))$ is equal to

Show that if $f: A \rightarrow B$ and $g: B \rightarrow C$ are onto,then $g \circ f: A \rightarrow C$ is also onto.

For every real number $x \neq -1$,let $f(x) = \frac{x}{x+1}$. Define $f_1(x) = f(x)$ and for $n \geq 2$,$f_n(x) = f(f_{n-1}(x))$. Then the product $f_1(-2) \cdot f_2(-2) \cdot \ldots \cdot f_n(-2)$ is equal to:

Let $f(x) = e^x$ and $g(x) = x^2$. Then,the number of solutions of $f(g(x)) = g(f(x))$ is equal to:

Let the functions $f:(-1,1) \rightarrow R$ and $g:(-1,1) \rightarrow(-1,1)$ be defined by $f(x)=|2 x-1|+|2 x+1|$ and $g(x)=x-[x]$,where $[x]$ denotes the greatest integer less than or equal to $x$. Let $f \circ g:(-1,1) \rightarrow R$ be the composite function defined by $(f \circ g)(x)=f(g(x))$. Suppose $c$ is the number of points in the interval $(-1,1)$ at which $f \circ g$ is $NOT$ continuous,and suppose $d$ is the number of points in the interval $(-1,1)$ at which $f \circ g$ is $NOT$ differentiable. Then the value of $c+d$ is.