Question

Which one among the following statements is true for the function \(f(x)=x|x|\)?

• A. \(f(x)\) is discontinuous but not differentiable at \(x=0\).
• B. \(f(x)\) is discontinuous and differentiable at \(x=0\).
• C. \(f(x)\) is continuous but not differentiable at \(x=0\).
• D. \(f(x)\) is continuous and differentiable at \(x=0\).

Hint:

Write \(f(x)=x|x|\) using the definition of \(|x|\) as a two-piece function, then check the one-sided limits of \(f\) and of the difference quotient at \(x=0\) separately.

Answer 1

The Correct Option is D

Step 1: Write \(f(x)\) as a piecewise function.
By definition, \(|x|=x\) for \(x\geq0\) and \(|x|=-x\) for \(x<0\). So
\[ f(x)=x|x|= \begin{cases} x^2, & x\geq0\\ -x^2, & x<0 \end{cases} \] Both pieces are ordinary polynomials in \(x\), so away from \(x=0\) the function is clearly continuous and differentiable; the only point that needs checking is \(x=0\), where the two pieces join.

Step 2: Check continuity at \(x=0\).
The left-hand limit is
\[ \lim_{x\to0^-}f(x)=\lim_{x\to0^-}(-x^2)=0 \] The right-hand limit is
\[ \lim_{x\to0^+}f(x)=\lim_{x\to0^+}x^2=0 \] Both limits equal \(f(0)=0\), so \(f\) is continuous at \(x=0\).

Step 3: Check differentiability at \(x=0\) using the definition.
The derivative at \(0\), if it exists, is
\[ f'(0)=\lim_{h\to0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to0}\frac{h|h|-0}{h}=\lim_{h\to0}|h| \] Since \(|h|\to0\) as \(h\to0\) regardless of whether \(h\) approaches from the left or the right, this limit exists and equals \(0\). So \(f'(0)=0\), and \(f\) is differentiable at \(x=0\).

Step 4: Cross-check with the piecewise derivative.
For \(x>0\), \(f'(x)=2x\), which tends to \(0\) as \(x\to0^+\). For \(x<0\), \(f'(x)=-2x\), which also tends to \(0\) as \(x\to0^-\). Both one-sided derivatives agree at \(0\), confirming \(f'(0)=0\); in fact \(f(x)=x|x|\) is differentiable everywhere, with \(f'(x)=2|x|\).

Step 5: Eliminate the other options.

(A) and (B) claim discontinuity at \(x=0\): Step 2 showed the left and right limits both equal \(f(0)\), so \(f\) is continuous there; both are incorrect.

(C) claims continuous but not differentiable: Step 3 showed the derivative from the definition exists and equals \(0\), so \(f\) is in fact differentiable at \(x=0\); incorrect.

Final Answer:
The function \(f(x)=x|x|\) is continuous and differentiable at \(x=0\).
\[ \boxed{\text{Continuous and differentiable at } x=0} \]