Q no. 2
Question
Examples: Construct examples of the thing(s) described in the following. Try to find examples that are different than any in the reading.
(a) The graph of a function that is continuous, but not differentiable, at .
(b) The graph of a function that is left and right differentiable, but not differentiable, at .
(c) The graph of a function that is differentiable on the interval but not differentiable at the point
Step-by-Step Solution
Verified(a). $$
\begin{aligned}
f^{\prime}(2) &=\lim _{h \rightarrow 0^{+}} \frac{f(2+h)-f(-2)}{h} \\
&=-1
\end{aligned}
$$
(b). $\begin{aligned} f^{\prime}(3) &=\lim _{h \rightarrow 0^{+}} \frac{f(3+h)-f(3)}{h} \\ &=-1 \end{aligned}$
(c). $\begin{aligned} f^{\prime}(1) &=\lim _{h \rightarrow 0^{-}} \frac{f(1+h)-f(1)}{h} \\ &=-1 \end{aligned}$
Here the objective is to find the graph of a function that is continuous but not differentiable at $x=2$.
Consider a function:
$$
f(x)= \begin{cases}x-2, & \text { if } x \geq 2 \\ -x+2, & \text { if } x<2\end{cases}
$$
The graph of the function will be:
From the graph it can be obacrved that the function is continuous at $x=2$ but it is not differentiable at $x=2$ as
$$
\begin{aligned}
f^{\prime}(2) &=\lim _{h \rightarrow 0^{+}} \frac{f(2+h)-f(-2)}{h} \\
&=1
\end{aligned}
$$
But
$$
\begin{aligned}
f^{\prime}(2) &=\lim _{h \rightarrow 0^{+}} \frac{f(2+h)-f(-2)}{h} \\
&=-1
\end{aligned}
$$
Here the objective is to find the graph of a function that is left and right differentiable, but not differentiable at $x=3$.
Consider a function:
$$
f(x)= \begin{cases}x-3, & \text { if } x \geq 3 \\ -x+3, & \text { if } x<3\end{cases}
$$
The graph of the function will be:
From the graph it can be observed that the function is differentiable at any point $c$ such that $c<3$ and $c>3$ but it is not differentiable at $x=3$ as
$$
\begin{aligned}
f^{\prime}(3) &=\lim _{h \rightarrow 0^{\circ}} \frac{f(3+h)-f(3)}{h} \\
&=1
\end{aligned}
$$
But
$$
\begin{aligned}
f^{\prime}(3) &=\lim _{h \rightarrow 0^{-}} \frac{f(3+h)-f(3)}{h} \\
&=-1
\end{aligned}
$$
Here the objective is to find the graph of a function that is differentiable on the interval $[-1,1]$ but not differentiable at the point $x=1$.
Consider a function:
$$
f(x)= \begin{cases}x-1, & \text { if } x \geq 1 \\ -x+1, & \text { if } x<1\end{cases}
$$
The graph of the function will be:
From the graph it can be observed that the function is differentiable at any point $c$ such that $-1 \leq c<1$ but it is not differentiable at $x=1$ as
$$
\begin{aligned}
f^{\prime}(1) &=\lim _{h \rightarrow 0^{+}} \frac{f(1+h)-f(1)}{h} \\
&=1
\end{aligned}
$$
But
$$
\begin{aligned}
f^{\prime}(1) &=\lim _{h \rightarrow 0^{-}} \frac{f(1+h)-f(1)}{h} \\
&=-1
\end{aligned}
$$