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: Work with the compact form.
From \(f(x) = -\left(x - \dfrac{|x|}{2}\right)^2\), define \(g(x) = x - \dfrac{|x|}{2}\), so \(f(x) = -\,g(x)^2\). Note \(g(x) = \dfrac{x}{2}\) for \(x\ge0\) and \(g(x)=\dfrac{3x}{2}\) for \(x<0\); \(g\) is continuous at \(0\) (both give \(g(0)=0\)) but has a slope "kink" there (slope \(1/2\) on the right, \(3/2\) on the left), so \(g\) itself behaves like a scaled, asymmetric \(|x|\).

Step 2: Reason about \(f=-g^2\) using the chain rule.
Wherever \(g\) is differentiable (i.e. \(x\neq0\)), \(f'(x) = -2g(x)g'(x)\). For \(x>0\): \(f'(x) = -2\cdot\frac{x}{2}\cdot\frac12 = -\frac{x}{2}\). For \(x<0\): \(f'(x)=-2\cdot\frac{3x}{2}\cdot\frac32=-\frac{9x}{2}\). This matches the direct piecewise computation.

Step 3: Check continuity of \(f'\) at 0 using nearby sample points.
At \(x=0.01\): \(f'(0.01) = -0.005\). At \(x=0.0001\): \(f'(0.0001) = -0.00005\), clearly trending to \(0\) as \(x\to0^+\).
At \(x=-0.01\): \(f'(-0.01) = -\dfrac{9(-0.01)}{2} = 0.045\). At \(x=-0.0001\): \(f'(-0.0001) = 0.00045\), clearly trending to \(0\) as \(x\to0^-\) as well (the values shrink towards zero as \(x\) approaches zero from either side, even though the two branches have different slopes).
Both one-sided trends converge to \(0=f'(0)\), confirming \(f'\) is continuous at \(x=0\) (Option C true).

Step 4: Check the derivative of \(f'\) (second derivative) on both sides using the same sample points.
Slope of \(f'\) just to the right of 0: \(\dfrac{f'(0.01)-f'(0)}{0.01} = \dfrac{-0.005-0}{0.01}=-0.5\). Slope of \(f'\) just to the left of 0: \(\dfrac{f'(-0.01)-f'(0)}{-0.01} = \dfrac{0.045-0}{-0.01} = -4.5\). These two slopes (\(-0.5\) versus \(-4.5\)) disagree, confirming \(f'\) has a genuine corner at \(x=0\) and is therefore not differentiable there (Option D true).

Step 5: Maximum/minimum via the compact form.
Since \(f=-g^2 \le 0\) always, with equality only at \(x=0\), \(x=0\) is the unique global maximum (Option A true), and since \(f \to -\infty\) monotonically away from \(0\) on each side with no interior turning point, there is no local minimum (Option B false).

Step 6: Conclusion.
Consistent with the direct piecewise computation: true statements are (A), (C), (D); (B) is false. \[ \boxed{\text{(A), (C), (D)}} \]