For the function \( f(x)=\min\{|x|,|x-2|\} \) which of the following are TRUE?
(A) Function \( f(x) \) is continuous on R
(B) Function \( f(x) \) is differentiable on R
(C) Function \( f(x) \) is differentiable on \( \mathrm{R}\sim\{0,1,2\} \)
(D) Function is continuous on R but not differentiable at \( x=\displaystyle\frac{1}{2} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (C) only
|x| ≤ |x − 2| exactly when x is at least as close to 0 as to 2, that is when x ≤ 1. So
\( \displaystyle f(x)=\begin{cases}-x, & x\lt 0\\x, & 0\le x\lt 1\\2-x, & 1\le x\lt 2\\x-2, & x\ge 2\end{cases} \)
The slopes on the four pieces are −1, 1, −1, 1, and the pieces meet at x = 0, 1, 2 with the values 0, 1, 0.
So f is continuous on R, and it has corners (no derivative) exactly at x = 0, 1 and 2.
Checking the statements
(A) Function \( f(x) \) is continuous on R — Correct. The minimum of the two continuous functions |x| and |x − 2| is continuous.
(B) Function \( f(x) \) is differentiable on R — Incorrect. At x = 0, 1 and 2 the slope changes between −1 and 1.
(C) Function \( f(x) \) is differentiable on \( \mathrm{R}\sim\{0,1,2\} \) — Correct. With 0, 1 and 2 removed, f is linear on each piece.
(D) Function is continuous on R but not differentiable at \( x=\frac{1}{2} \) — Incorrect. Near \( x=\frac{1}{2} \), f(x) = x, so \( f'\left(\frac{1}{2}\right)=1 \) exists.
Tip: For the min or max of two functions, the point where the two graphs cross (here x = 1) is usually a point of non-differentiability too, besides the corners of each function.