Which one among the following statements is true for the function \(f(x)=x|x|\)?
Show 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.
\(f(x)\) is discontinuous but not differentiable at \(x=0\).
\(f(x)\) is discontinuous and differentiable at \(x=0\).
\(f(x)\) is continuous but not differentiable at \(x=0\).
\(f(x)\) is continuous and differentiable at \(x=0\).
Show Solution
The Correct Option isD
Solution and Explanation
Step 1: Spot a shortcut form.
Since $|x|=\sqrt{x^2}$, we can write $f(x)=x\sqrt{x^2}$. This is built from continuous, differentiable pieces everywhere except possibly at the one point where $\sqrt{x^2}$ has a sharp corner in isolation, namely $x=0$.
Step 2: Test continuity by direct substitution.
As $x\to0$, both $x\to0$ and $|x|\to0$, so their product $x|x|\to0\times0=0$, matching $f(0)=0$. There is no jump, so $f$ is continuous at $x=0$.
Step 3: Bound the difference quotient instead of splitting into cases.
The difference quotient at $0$ is
\[ \frac{f(h)-f(0)}{h}=\frac{h|h|}{h}=|h| \]
for every $h\neq0$; this single expression already covers both $h>0$ and $h<0$, since $h/h=1$ in both cases. As $h\to0$, $|h|\to0$, so the limit of the difference quotient exists and equals $0$. Hence $f'(0)=0$.
Step 4: Confirm smoothness away from zero, then combine.
For $x\neq0$, $f(x)$ is either $x^2$ or $-x^2$, both polynomials, so continuous and differentiable there with derivative $2|x|$. Combined with Step 3, $f$ is continuous and differentiable at every real $x$, including $x=0$.
Step 5: Match to the options.
Since $f$ is neither discontinuous (rules out A and B) nor non-differentiable (rules out C) at $x=0$, the only statement left standing says it is continuous and differentiable there.
\[ \boxed{\text{Continuous and differentiable at } x=0} \]
