Given the function $h(r)=\left\{\begin{array}{cc}1-\sin (\pi r / 2) & -1 \leq r \leq 1 \\0 & r<-1, r>1\end{array}\right.$ .Is $h(r)$
differentiable at r = -1?

Question

Given the function $h(r)=\left\{\begin{array}{cc}
1-\sin (\pi r / 2) & -1 \leq r \leq 1 \
0 & r<-1, r>1
\end{array}\right.$ .Is $h(r)$
 differentiable at r = -1?

Quizplus · Accepted Answer

The function $h(r)$ involves a piecewise definition with a transition at $r = -1$. To be differentiable at $r = -1$, $h(r)$ must be continuous at that point, and the left-hand derivative and the right-hand derivative at $r = -1$ must be equal. While $h(r)$ is continuous at $r = -1$, the derivative from the left does not exist because the function is constant (and thus the derivative is 0) for $r < -1$, and the derivative from the right involves the derivative of $1 - \sin(\pi r / 2)$, which is not zero. Therefore, the derivatives from the left and right do not match, making $h(r)$ not differentiable at $r = -1$.

Given the Function h(r)={1−sin⁡(πr/2)−1≤r≤10r<−1,r>1h(r)=\left\{\begin{array}{cc}1-\sin (\pi r / 2) & -1 \leq r \leq 1 \\0 & r<-1, r>1\end{array}\right.h(r)={1−sin(πr/2)0​−1≤r≤1r<−1,r>1​

Given the Function $h(r)=\left\{\begin{array}{cc}1-\sin (\pi r / 2) & -1 \leq r \leq 1 \\0 & r<-1, r>1\end{array}\right.$