Question

Let \(f: \mathbb{R} \to \mathbb{R}\) be defined as follows: \[ f(x) = \left(\dfrac{|x|}{2} - x\right)\left(x - \dfrac{|x|}{2}\right) \] 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 \(\mathbb{R}\)
• D. \(f'\) is not differentiable over \(\mathbb{R}\)

Hint:

Simplify f(x) to -(x-|x|/2)^2 first, then examine the piecewise quadratic branches for extrema, and check the one-sided limits of f' and of f'' separately at x=0.

Answer 1

The Correct Option is A, C, D

Step 1: Simplify \(f(x)\) algebraically.
Notice that \(\dfrac{|x|}{2} - x = -\left(x - \dfrac{|x|}{2}\right)\). So \[ f(x) = \left(\dfrac{|x|}{2}-x\right)\left(x-\dfrac{|x|}{2}\right) = -\left(x - \dfrac{|x|}{2}\right)^2 \] This is always \(\le 0\), and equals \(0\) exactly when \(x = \dfrac{|x|}{2}\), i.e. only at \(x=0\).

Step 2: Write \(f(x)\) piecewise.
For \(x \ge 0\): \(|x|=x\), so \(x - \dfrac{|x|}{2} = x - \dfrac{x}{2} = \dfrac{x}{2}\), giving \(f(x) = -\left(\dfrac{x}{2}\right)^2 = -\dfrac{x^2}{4}\).
For \(x < 0\): \(|x|=-x\), so \(x - \dfrac{|x|}{2} = x-\left(\dfrac{-x}{2}\right) = x + \dfrac{x}{2} = \dfrac{3x}{2}\), giving \(f(x) = -\left(\dfrac{3x}{2}\right)^2 = -\dfrac{9x^2}{4}\).
So: \[ f(x) = \begin{cases} -\dfrac{x^2}{4}, & x \ge 0 \\[4pt] -\dfrac{9x^2}{4}, & x < 0 \end{cases} \] Both branches are downward-opening parabolic pieces meeting at \(f(0)=0\), and \(f(x) < 0\) for all \(x \neq 0\).

Step 3: Examine local maxima/minima (Options A and B).
Since \(f(0)=0\) and \(f(x)<0\) everywhere else, \(x=0\) is the strict global maximum, hence also a local maximum: Option (A) is TRUE.
On \((0,\infty)\), \(f\) is strictly decreasing (\(-x^2/4\) decreases as \(x\) increases from 0); on \((-\infty,0)\), \(f\) is strictly increasing towards 0 (\(-9x^2/4\) increases as \(x\) increases towards 0). There is no interior point where \(f\) turns from decreasing to increasing, and \(f \to -\infty\) as \(x \to \pm\infty\), so \(f\) has no local minimum anywhere: Option (B) is FALSE.

Step 4: Differentiate piecewise.
For \(x>0\): \(f'(x) = -\dfrac{x}{2}\).
For \(x<0\): \(f'(x) = -\dfrac{9x}{2}\).
At \(x=0\), compute directly from the definition of the derivative: \[ f'(0) = \lim_{h \to 0} \frac{f(h)-f(0)}{h} \] From the right (\(h\to0^+\)): \(\dfrac{-h^2/4 - 0}{h} = -\dfrac{h}{4} \to 0\).
From the left (\(h\to0^-\)): \(\dfrac{-9h^2/4-0}{h} = -\dfrac{9h}{4} \to 0\).
Both one-sided limits agree, so \(f'(0)=0\) exists, and: \[ f'(x) = \begin{cases} -\dfrac{x}{2}, & x \ge 0 \\[4pt] -\dfrac{9x}{2}, & x < 0 \end{cases} \]
Step 5: Check continuity of \(f'\) (Option C).
\(\displaystyle\lim_{x\to0^+} f'(x) = \lim_{x\to0^+}\left(-\dfrac{x}{2}\right) = 0\), and \(\displaystyle\lim_{x\to0^-} f'(x) = \lim_{x\to0^-}\left(-\dfrac{9x}{2}\right)=0\). Both one-sided limits equal \(f'(0)=0\). Away from \(x=0\), \(f'\) is a simple linear function on each piece, hence continuous there too. So \(f'\) is continuous at every point of \(\mathbb{R}\): Option (C) is TRUE.

Step 6: Check differentiability of \(f'\) (Option D).
Differentiate \(f'\) again where possible: for \(x>0\), \((f')'(x) = -\dfrac12\); for \(x<0\), \((f')'(x)=-\dfrac92\). Taking one-sided derivatives of \(f'\) at \(x=0\): the right-hand derivative of \(f'\) at 0 is \(-\dfrac12\), the left-hand derivative is \(-\dfrac92\). These are unequal, so \(f'\) is NOT differentiable at \(x=0\) (it has a corner there, similar to \(|x|\), even though \(f'\) itself is continuous). Since \(f'\) fails to be differentiable at at least one point of \(\mathbb{R}\), the claim "\(f'\) is differentiable over \(\mathbb{R}\)" is false, so its negation, "\(f'\) is not differentiable over \(\mathbb{R}\)", is TRUE: Option (D) is TRUE.

Step 7: Conclusion.
True statements: (A) local maximum, (C) \(f'\) continuous over \(\mathbb{R}\), (D) \(f'\) not differentiable over \(\mathbb{R}\). False: (B). \[ \boxed{\text{(A), (C), (D)}} \]