Question

If
\[ f(x)=\max\{3-x,\;3+x,\;6\} \] is not differentiable at \(x=a\), and \(x=b\), then \(|a|+|b|=\)

• A. \(4\)
• B. \(5\)
• C. \(6\)
• D. \(8\)

Hint:

For functions involving \(\max\), non-differentiability usually occurs at points where the maximum switches from one expression to another.

Answer 1

The Correct Option is C

Step 1: List the three competing pieces.
The function is $f(x)=\max\{3-x,\,3+x,\,6\}$, the upper envelope of the line $3-x$, the line $3+x$ and the constant $6$.
Step 2: Find where the slanted lines meet the constant.
$3-x=6$ gives $x=-3$, and $3+x=6$ gives $x=3$. These are the candidate corner points.
Step 3: Identify the winner on the left.
For $x<-3$, $3-x>6$, so $f(x)=3-x$ (slope $-1$).
Step 4: Identify the winner in the middle.
For $-3<x<3$, both lines are below $6$, so $f(x)=6$ (slope $0$).
Step 5: Identify the winner on the right.
For $x>3$, $3+x>6$, so $f(x)=3+x$ (slope $+1$). The slope jumps at $x=-3$ (from $-1$ to $0$) and at $x=3$ (from $0$ to $1$), so $f$ is non differentiable at $a=-3$ and $b=3$.
Step 6: Add the absolute values.
$|a|+|b|=|-3|+|3|=6$.
\[ \boxed{6} \]