Question

If \[f(x)=\left\{ \begin{matrix}
   {{x}^{3}}+1,x<0 \\
   {{x}^{2}}+1,x\ge 0 \\
\end{matrix},g(x)=\left\{ \begin{matrix}
   {{(x-1)}^{1/3}},x<1 \\
   {{(x-1)}^{1/3}},x\ge 1 \\
\end{matrix}, \right. \right.\]then $(gof)(x)$ is equal to
A. $x,\forall \,\,x\in R$
B. $x-1,\forall x\in R$
C. $x+1,\forall x\in R$
D. None of these.

Answer 1

Hint: To solve this question, we will consider two cases for the value of $x$ and then substitute the value in the function, value of which we have to derive. We will consider $x<0$ and $x\ge 0$ and determine the value of $(gof)(x)$ by substituting the value of functions $f(x)$ and $g(x)$ accordingly.

Complete step by step solution: We are given two piecewise functions \[f(x)=\left\{ \begin{matrix}
   {{x}^{3}}+1,x<0 \\
   {{x}^{2}}+1,x\ge 0 \\
\end{matrix},g(x)=\left\{ \begin{matrix}
   {{(x-1)}^{1/3}},x<1 \\
   {{(x-1)}^{1/3}},x\ge 1 \\
\end{matrix}, \right. \right.\] and we have to determine the value of $(gof)(x)$.
We will first rewrite the equation $(gof)(x)$ as $g(f(x))$and then substitute the value of both the functions $f(x)$ and $g(x)$.
We will first consider a value of $x$ less than zero that is $x<0$ and determine the value of $(gof)(x)$.
As $x<0$, we will take value of the function $f(x)={{x}^{3}}+1$ and $g(x)={{(x-1)}^{\dfrac{1}{3}}}$.
$\begin{align}
  & (gof)(x)=g(f(x)) \\
 & =g({{x}^{3}}+1) \\
 & ={{\left[ ({{x}^{3}}+1)-1 \right]}^{\dfrac{1}{3}}}
\end{align}$
Because the value of $x$ is less than zero, then the value ${{x}^{3}}+1<1$. So,
$\begin{align}
  & (gof)(x)={{\left[ {{x}^{3}} \right]}^{\dfrac{1}{3}}} \\
 & =x
\end{align}$
We will now consider the value of$x$greater than zero that is $x\ge 0$.
As $x\ge 0$, we will take the value of function $f(x)={{x}^{2}}+1$ and $g(x)={{(x-1)}^{\dfrac{1}{2}}}$.
$\begin{align}
  & (gof)(x)=g(f(x)) \\
 & =g({{x}^{2}}+1) \\
 & ={{\left[ ({{x}^{2}}+1)-1 \right]}^{\dfrac{1}{2}}}
\end{align}$
Because the value of $x$ is greater than zero, then the value ${{x}^{2}}+1\le 1$. So,
$\begin{align}
  & (gof)(x)={{\left[ {{x}^{2}} \right]}^{\dfrac{1}{2}}} \\
 & =x
\end{align}$
The values for both the cases are similar hence we can say that the value of $(gof)(x)$ for all the values will be a real number that is $x,\forall \,\,x\in R$.
The value of function$(gof)(x)$ when\[f(x)=\left\{ \begin{matrix}
   {{x}^{3}}+1,x<0 \\
   {{x}^{2}}+1,x\ge 0 \\
\end{matrix},g(x)=\left\{ \begin{matrix}
   {{(x-1)}^{1/3}},x<1 \\
   {{(x-1)}^{1/3}},x\ge 1 \\
\end{matrix}, \right. \right.\] is $x,\forall \,\,x\in R$.

So, Option ‘A’ is correct

Note: We can state that a piecewise function is a type of function which is defined by many sub functions and each of that sub function is defined on a different interval. The functions we are given have two sub functions for both the functions $f(x)$ and $g(x)$ on two intervals.