Question

Let \[ f(x)= \begin{cases} 4, & -\infty\lt x\lt -\sqrt5 \\ x^2-1, & -\sqrt5\leq x\leq \sqrt5 \\ 4, & \sqrt5\lt x\lt \infty \end{cases} \] If \(k\) is the number of points where \(f(x)\) is not differentiable, then \(k-2=\)

• A. \(2\)
• B. \(1\)
• C. \(0\)
• D. \(3\)

Hint:

For a piecewise function, possible non-differentiability occurs at the joining points. Check the left hand derivative and right hand derivative at each joining point.

Answer 1

The Correct Option is C

Step 1: Identify possible points of non-differentiability.
The function is defined in different pieces. Therefore, possible points of non-differentiability occur at the joining points: \[ x=-\sqrt5 \] and \[ x=\sqrt5. \] For all other intervals, the function is either constant or polynomial, so it is differentiable there.

Step 2: Check differentiability at \(x=-\sqrt5\).
For \[ x\lt -\sqrt5, \] \[ f(x)=4. \] So, the left hand derivative is \[ f'_-\left(-\sqrt5\right)=0. \] For \[ -\sqrt5\leq x\leq \sqrt5, \] \[ f(x)=x^2-1. \] So, \[ f'(x)=2x. \] Hence, the right hand derivative at \[ x=-\sqrt5 \] is \[ f'_+\left(-\sqrt5\right)=2(-\sqrt5)=-2\sqrt5. \] Since \[ 0\neq -2\sqrt5, \] \(f(x)\) is not differentiable at \[ x=-\sqrt5. \]

Step 3: Check differentiability at \(x=\sqrt5\).
For \[ -\sqrt5\leq x\leq \sqrt5, \] \[ f(x)=x^2-1. \] So, \[ f'(x)=2x. \] Hence, the left hand derivative at \[ x=\sqrt5 \] is \[ f'_-\left(\sqrt5\right)=2\sqrt5. \] For \[ x\gt \sqrt5, \] \[ f(x)=4. \] So, the right hand derivative is \[ f'_+\left(\sqrt5\right)=0. \] Since \[ 2\sqrt5\neq 0, \] \(f(x)\) is not differentiable at \[ x=\sqrt5. \]

Step 4: Find the value of \(k\).
The function is not differentiable at two points: \[ x=-\sqrt5 \] and \[ x=\sqrt5. \] Therefore, \[ k=2. \] Now, \[ k-2=2-2=0. \]

Step 5: Final conclusion.
Therefore, \[ \boxed{0} \]