If f(x)={−x−π2,x≤−π2−cos⁡x,−π2<x≤0x−1,0<x≤1ln⁡x,x>1f(x) = \begin{cases} -x - \frac{\pi}{2}, & x \le -\frac{\pi}{2} \\ -\cos x, & -\frac{\pi}{2} < x \le 0 \\ x - 1, & 0 < x \le 1 \\ \ln x, & x > 1 \end{cases}…

Mathematics

Problem

If f(x)={−x−π2,x≤−π2−cos⁡x,−π2<x≤0x−1,0<x≤1ln⁡x,x>1f(x) = \begin{cases} -x - \frac{\pi}{2}, & x \le -\frac{\pi}{2} \\ -\cos x, & -\frac{\pi}{2} < x \le 0 \\ x - 1, & 0 < x \le 1 \\ \ln x, & x > 1 \end{cases} a. f(x)f(x) is continuous at x=−π2x = -\frac{\pi}{2} b. f(x)f(x) is not differentiable at x=0x = 0 c. f(x)f(x) is differentiable at x=1x = 1 d. f(x)f(x) is differentiable at x=−32x = -\frac{3}{2}

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