Question

Let \(f:\mathbb{R}\to\mathbb{R}\) be defined by
\[f(x)=\left(\frac{|x|}{2}-x\right)\left(x-\frac{|x|}{2}\right).\]
Which of the following statements is/are true?

• A. 𝑓 has a local maximum
• B. 𝑓 has a local minimum
• C. 𝑓′ is continuous over ℝ
• D. 𝑓′ is not differentiable over ℝ

Hint:

Simplify \( f(x) = -\left(x - \frac{|x|}{2}\right)^2 \), split into cases \( x \ge 0 \) (giving \( -x^2/4 \)) and \( x < 0 \) (giving \( -9x^2/4 \)). Check the sign of \( f \) near 0 for max/min, then check whether \( f' \) matches on both sides (continuity) and whether the slopes of \( f' \) match on both sides (differentiability of \( f' \)).

Answer 1

The Correct Option is A, C, D

Step 1: Simplify the function. We are given \[ f(x) = \left(\frac{|x|}{2} - x\right)\left(x - \frac{|x|}{2}\right) \] Let \( g(x) = x - \frac{|x|}{2} \). Then the first factor is \( -g(x) \), so \[ f(x) = -g(x)\cdot g(x) = -\left(x - \frac{|x|}{2}\right)^2 \] This means \( f(x) \le 0 \) for every real \( x \), with equality only at \( x = 0 \).

Step 2: Write \( f \) piecewise using the definition of \( |x| \).

For \( x \ge 0 \): \( |x| = x \), so \[ f(x) = \left(\frac{x}{2} - x\right)\left(x - \frac{x}{2}\right) = \left(-\frac{x}{2}\right)\left(\frac{x}{2}\right) = -\frac{x^2}{4} \]

For \( x < 0 \): \( |x| = -x \), so \[ f(x) = \left(-\frac{x}{2} - x\right)\left(x + \frac{x}{2}\right) = \left(-\frac{3x}{2}\right)\left(\frac{3x}{2}\right) = -\frac{9x^2}{4} \]

Step 3: Check local maximum/minimum (Options A and B). At \( x = 0 \), \( f(0) = 0 \). For any \( x \ne 0 \) near 0, \( f(x) < 0 \) on both sides (since both branches are negative parabolas opening downward). So \( f(0) = 0 \) is greater than all nearby values, meaning \( x = 0 \) is a local maximum (in fact a global maximum). So Option A is true.

Since both branches \( -\frac{x^2}{4} \) and \( -\frac{9x^2}{4} \) decrease without bound as \( |x| \to \infty \) and there is no interior point where the function turns from decreasing to increasing, \( f \) has no local minimum anywhere on \( \mathbb{R} \). So Option B is false.

Step 4: Check continuity of \( f' \) (Option C). Differentiating each branch: \[ f'(x) = \begin{cases} -\dfrac{x}{2}, & x \ge 0 \\ -\dfrac{9x}{2}, & x < 0 \end{cases} \] As \( x \to 0^+ \), \( f'(x) \to 0 \). As \( x \to 0^- \), \( f'(x) \to 0 \). Both one-sided limits equal \( f'(0) = 0 \) (verified by the difference quotient of \( f \) at 0, which also gives 0 from both sides). Away from 0, \( f' \) is a polynomial on each side, hence continuous. So \( f' \) is continuous everywhere on \( \mathbb{R} \). Option C is true.

Step 5: Check differentiability of \( f' \) at 0 (Option D). Compute one-sided derivatives of \( f' \) at \( x = 0 \): \[ \lim_{x\to 0^+} \frac{f'(x)-f'(0)}{x} = \lim_{x\to 0^+}\frac{-x/2}{x} = -\frac{1}{2} \] \[ \lim_{x\to 0^-} \frac{f'(x)-f'(0)}{x} = \lim_{x\to 0^-}\frac{-9x/2}{x} = -\frac{9}{2} \] Since \( -\frac{1}{2} \ne -\frac{9}{2} \), \( f' \) has a corner (kink) at \( x = 0 \) and is not differentiable there, even though it is continuous. So Option D is true.

Final Answer:

The correct options are A, C and D.