Q. 2 - Calculus

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)  A function with a local minimum at x = 3 that is continuous but not differentiable at x = 3 .

(b)  A function with a local maximum at x = −2 that is not differentiable at x = −2 because of a removable discontinuity .

(c)  A function with a local minimum at x = 1 that is not differentiable at x = 1 because of a jump discontinuity .

Step-by-Step Solution

Answer

(a) Function is continuous but not differentiable .

(b) Function is not differentiable because or removable discontinuity .

(c) Function is not differentiable because of jump discontinuity .

Step 1. (a) A function with a local minimum at x = 3 that is continuous but not differentiable at x = 3.

We can write $f\left(x\right)$ as .

$f\left(x\right)$ $=x-3 , x\ge 3$

          $=3-x , x<3$

LHL for $f\left(x\right)$ at $x=3$is $0$ .

RHL for $f\left(x\right)$ at $x=3$ is $0$ .

So $f\left(x\right)$ is continuous at $x=3$ .

Further simplify .

RHD of $f\left(x\right)$ at $x=3$ $\underset{h\to 0}{\lim }\left\{\frac{\left(3+h\right)-3-\left(3-3\right)}{3+h-3}\right\}=1$

LHD of $f\left(x\right)$ at $x=3$ $\underset{h\to 0}{\lim }\left\{\frac{3-\left(3-h\right)-\left(3-3\right)}{3-h-3}\right\}=-1$

So $f\left(x\right)$ is not differentiable at $x=3$ .

Step 2. A function with a local maximum at x = &minus;2 that is not differentiable at x = &minus;2 because of a removable discontinuity .

 If the function is not smooth and continuous it is not differentiable at least over all real numbers. For asymptotic functions and functions with point discontinuities, the derivative only doesn’t exist at a point, which still counts as differentiable .

Step 3. A function with a local minimum at x = 1 that is not differentiable at x = 1 because of a jump discontinuity.

Determine the type of discontinuity at  $x=0$ for $f\left(x\right)=\frac{x}{\left|x\right|}$.

$$\begin{gathered} \left|x\right|=\left\{x , x>0 , -x , x<0\right\} \\ LHS =\underset{x\to 0}{\lim }\frac{x}{\left|x\right|}=-1 \\ RHS =\underset{x\to 0}{\lim }\frac{x}{\left|x\right|}=1 \end{gathered}$$

Since LHS is not equal to RHS so there is jump discontinuity .