Question

Let \[ f(x)= \begin{cases} -150-25x, & -10\le x\le -5,[2mm] x|x|, & -5\le x\le 5,[2mm] 150-25x, & 5\le x\le 10. \end{cases} \] Which of the following is correct?

• A. \(f(x)\) is continuous at all points except at \(x=\pm5\)
• B. \(f(x)\) is differentiable at all points except at \(x=\pm5\)
• C. \(f(x)\) is not differentiable at \(x=0\) since \(|x|\) is not differentiable at \(x=0\)
• D. \(f(x)\) has \(4\) points of discontinuity

Hint:

A piecewise function may be continuous but not differentiable at the joining points. Always check both continuity and equality of the left and right derivatives.

Answer 1

The Correct Option is B

Step 1: Check continuity at the junction points. At \[ x=-5, \] \[ \lim_{x\to-5^-}f(x) =-150-25(-5) =-25, \] and \[ \lim_{x\to-5^+}f(x) =(-5)|-5| =-25. \] Also, \[ f(-5)=-25. \] Hence, \[ f(x) \] is continuous at \[ x=-5. \] Similarly, at \[ x=5, \] \[ \lim_{x\to5^-}f(x) =5|5| =25, \] \[ \lim_{x\to5^+}f(x) =150-25(5) =25, \] and \[ f(5)=25. \] Hence, \[ f(x) \] is also continuous at \[ x=5. \]

Step 2: Check differentiability. For \[ -5<x<0, \] \[ f(x)=-x^2, \] so \[ f'(x)=-2x. \] For \[ 0<x<5, \] \[ f(x)=x^2, \] so \[ f'(x)=2x. \] At \[ x=0, \] \[ \lim_{x\to0^-}f'(x)=0, \qquad \lim_{x\to0^+}f'(x)=0. \] Hence, \[ f(x) \] is differentiable at \[ x=0. \] At \[ x=-5, \] \[ f'_-( -5 )=-25, \qquad f'_+( -5 )=10. \] Thus, \[ f(x) \] is not differentiable at \[ x=-5. \] At \[ x=5, \] \[ f'_-(5)=10, \qquad f'_+(5)=-25. \] Hence, \[ f(x) \] is not differentiable at \[ x=5. \]

Step 3: Write the conclusion. Therefore, \[ f(x) \] is differentiable everywhere except at \[ x=\pm5. \] Hence, \[ \boxed{\text{\(f(x)\) is differentiable at all points except at }x=\pm5.} \] Thus, \[ \boxed{(B)} \] is the correct answer.