Question

Let f(x) = $\{\begin{array}{c}1+x^3,x<0\\ x^2-1,x\ge 0\end{array}$ ; g(x) = $\{\begin{array}{c}(x-1{)}^{1/3},x<0\\ (x+1{)}^{1/2},x\ge 0\end{array}$. Discuss the continuity of g(f(x))

• A. gof is discontinuous at x = 0
• B. gof is discontinuous at x = 1
• C. gof is discontinuous at x =-1
• D. gof is continuous at x = 0

Answer 1

Answer: (A), (B), (C)

The correct options are
A gof is discontinuous at x = 0
B gof is discontinuous at x =-1
C gof is discontinuous at x = 1

$f\left(x\right)=\{\begin{array}{c}1+x^{3}x<0\\ x^2-1x\ge 0\end{array}$

$g\left(x\right)=\{\begin{array}{c}(x-1{)}^{\frac{1}{3}}x<0\\ (x+1{)}^{1/2}x\ge 0\end{array}$

$g\left(f\right(x\left)\right)=\{\begin{array}{c}\left(f\right(x)-1{)}^{\frac{1}{3}}f(x)<0\\ \left(f\right(x)+1{)}^{1/2}f(x)\ge 0\end{array}$

$f\left(x\right)=\{\begin{array}{c}1+x^{3}x<-1\\ 1+x^3-1\le x\le 0\\ x^2-10<x<1\\ x^2-1x\ge 1\end{array}$

$g\left(f\right(x\left)\right)=\{\begin{array}{c}(1+x^3-1{)}^{\frac{1}{3}}x<-1\\ (1+x^3+1{)}^{1/2}-1\le x\le 0\\ (x^2-1-1{)}^{\frac{1}{3}}0<x<1\\ (x^2-1+1{)}^{1/2}x\ge 1\end{array}$

$\Rightarrow \underset{x\to {-1}^{-}}{lim}g\left(f\right(x\left)\right)=\underset{x\to {-1}^{-}}{lim}x=-1$

$\Rightarrow \underset{x\to {-1}^{+}}{lim}g\left(f\right(x\left)\right)=li{m}_{x\to {-1}^{-}}(1+x^3+1{)}^{1/2}=1$

$\Rightarrow \underset{x\to 0^{-}}{lim}g\left(f\right(x\left)\right)=\underset{x\to 0^{-}}{lim}(1+x^3+1{)}^{1/2}=\sqrt{2}$

$\Rightarrow li{m}_{x\to 0^{+}}g\left(f\right(x\left)\right)=\underset{x\to 0^{+}}{lim}(x^2-1-1{)}^{\frac{1}{3}}={(-2)}^{1/3}$

$\underset{x\to 1^{-}}{lim}g\left(f\right(x\left)\right)=\underset{x\to 1^{-}}{lim}(x^2-1-1{)}^{\frac{1}{3}}=-1$

$\Rightarrow \underset{x\to 1^{+}}{lim}g\left(f\right(x\left)\right)=\underset{x\to 1^{+}}{lim}x=1$

$g\left(f\right(x\left)\right)$ is discontinuous
at $x=-1,0,1$