Let 𝑓:ℝ→ℝ be defined as follows: f(𝑥)=(|𝑥|/2 −𝑥)(𝑥−|𝑥|/2 ) Which of the…

For x ≥ 0, |x| = x, so f(x) = (x/2 - x)(x - x/2) = (-x/2)(x/2) = -x²/4. For x < 0, |x| = -x, so f(x) = (-x/2 - x)(x + x/2) = (-3x/2)(3x/2) = -9x²/4. Thus f(0) = 0 and f(x) < 0 for x ≠ 0 near the origin, so f has a local maximum at 0 and not a local minimum. Also, f′(x) = -9x/2 for x < 0, f′(0) = 0, and f′(x) = -x/2 for x > 0, hence f′ is continuous at 0. However, the left derivative of f′ at 0 is -9/2 and the right derivative is -1/2, so f′ is not differentiable at 0. Therefore, statements 1, 3, and 4 are true.