Differentiability
Continuity
MMTS_Full_Test_02
Grade 12
Question:
If $g(x)=f(f(x))$ where $f(x)=\begin{cases}1+x & 0\le x\le1\\3-x & 1\le x\le 2\end{cases}$, then $g'(x)$ at $x=1$
doesn't exist
$1$
$0$
$-1$
Step-by-Step Solution
Key Concept: Find $g(x)=f(f(x))$ and check differentiability at $x=1$
For $x$ near 1: $f(x)=1+x$ for $x<1$: $g(x)=f(1+x)=3-(1+x)=2-x$. $g'=−1$ from left. For $x>1$: $f(x)=3-x$: $g(x)=f(3-x)$: $3-x\in[1,2]$ near $x=1$: $g(x)=3-(3-x)=x$. $g'=1$ from right. Doesn't exist.
Correct Answer: 2