Limits, Continuity & Differentiability
Differentiability of Composite Piecewise Function
nta_pyq_2023_apr
Grade 12
Question:
Let $f$ and $g$ be two functions defined by $f(x)=\begin{cases}x+1, & x<0\\|x-1|, & x\geq0\end{cases}$ and $g(x)=\begin{cases}x+1, & x<0\\1, & x\geq0\end{cases}$. Then $(g\circ f)(x)$ is
Continuous everywhere but not differentiable exactly at one point
Continuous everywhere but not differentiable at x=1
Differentiable everywhere
Not continuous at x=1
Step-by-Step Solution
Key Concept: Compute $g(f(x))$ case by case: for $x<-1$, $f(x)<0$ so $g(f(x))=f(x)+1=x+2$; for $x\geq-1$, $f(x)\geq0$ so $g(f(x))=1$.
$g\circ f(x)=x+2$ for $x<-1$, $=1$ for $x\geq-1$. Continuous, not differentiable at $x=-1$ only.
Correct Answer: 1