Question

The function \(f(x)=x|x|\) is

• A. not monotonic
• B. strictly decreasing function
• C. differentiable for all \(x\in R\)
• D. differentiable for all \(x\in R\) except at \(x=0\)

Hint:

For modulus functions, always check differentiability at the point where the expression inside modulus becomes zero.

Answer 1

The Correct Option is C

Concept:
A function involving modulus may or may not be differentiable at the point where the modulus changes sign. Here, \[ f(x)=x|x| \] We write it in piecewise form.

Step 1: Write \(f(x)\) for \(x\geq 0\). If \[ x\geq 0 \] then \[ |x|=x \] So, \[ f(x)=x\cdot x=x^2 \]

Step 2: Write \(f(x)\) for \(x<0\). If \[ x<0 \] then \[ |x|=-x \] So, \[ f(x)=x(-x)=-x^2 \] Thus, \[ f(x)= \begin{cases} -x^2, & x\\ x^2, & x\geq 0. \end{cases} \]

Step 3: Check differentiability for \(x\neq 0\). For \(x<0\), \[ f'(x)=-2x \] For \(x>0\), \[ f'(x)=2x \] So the function is differentiable for all \(x\neq 0\).

Step 4: Check differentiability at \(x=0\). Right derivative: \[ f'_+(0)=\lim_{h\to 0^+}\frac{f(h)-f(0)}{h} \] For \(h>0\), \[ f(h)=h^2 \] So, \[ f'_+(0)=\lim_{h\to 0^+}\frac{h^2}{h} \] \[ =\lim_{h\to 0^+}h=0 \] Left derivative: For \(h<0\), \[ f(h)=-h^2 \] So, \[ f'_-(0)=\lim_{h\to 0^-}\frac{-h^2}{h} \] \[ =\lim_{h\to 0^-}(-h)=0 \] Since left derivative and right derivative are equal, \[ f'(0)=0 \]

Step 5: Final answer. Therefore, \(f(x)=x|x|\) is differentiable for all real \(x\). \[ \boxed{\text{differentiable for all }x\in R} \]