Question

In Exercises \(31-36,\) find all values of \(x\) for which the function is differentiable. $$P(x)=\sin (|x|)-1$$

Short answer

The function \(P(x)=\sin (|x|)-1\) is differentiable for all \(x \in \mathbb{R}\), but not at \(x = 0\)

Step-by-step solution

Differentiation of the Function

The function \(P(x)=\sin (|x|)-1\) is not differentiable at \(x = 0\). We will have to break down the function into two separate functions for \(x < 0\) and \(x > 0\), and differentiate both separately. For \(x > 0\), the function simplifies as \(P(x)=\sin (x)-1\) and for \(x < 0\), it simplifies as \(P(x)=\sin (-x)-1\), then differentiate these functions.

Differentiation for \(x > 0\) and \(x < 0\)

Derivate both functions obtained from step 1. For \(x > 0\), \(P'(x) = \cos(x)\); for \(x < 0\), \(P'(x) = -\cos(-x) = -\cos(x)\). However, as explained before, the original function is not differentiable at \(x = 0\), therefore, the domain of differentiability of the given function will be all real numbers except 0.

Summarize the Differentiability of the Given Function

After checking the differentiability for \(x < 0\) and \(x > 0\), and keeping in mind that the function is not differentiable at \(x = 0\), we can conclude that the function \(P(x)=\sin (|x|)-1\) is differentiable for all real numbers \(x\) excluding \(x = 0\)