Q. 2 - Calculus

Question

Q. 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 with a local minimum at x = 2 but no global minimum on [0, 4].

(b) The graph of a function with no local or global extrema on (−3, 3).

(c) The graph of a function whose global maximum on [2, 6] does not occur at a critical point.

Step-by-Step Solution

Answer

(a) The graph of a function with a local minimum at x = 2 but no global minimum on [0, 4] is:

(b) The graph of a function with no local or global extrema on (−3, 3) is:

(c) The graph of a function whose global maximum on [2, 6] does not occur at a critical point is:

Part (a) Step 1. Given Information.

A function with a local minimum at x = 2 but no global minimum on [0, 4]. 

Part (a) Step 2. Find the function.

Consider the function,

$$\begin{gathered} f\left(x\right)=\frac{x^2}{{(x-1)}^2} \\ f\text{'}\left(x\right)=\frac{x(x-2)}{{(x-1)}^2}=0 \\ x=0,2 \end{gathered}$$

The critical points are $x=0,2$

Part (a) Step 3. Find the second derivative.

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{2}{{(x-1)}^3} \\ f\text{'}\text{'}\left(2\right)=2>0 \\ f\left(2\right)=4 \end{gathered}$$

The function is not continuous at x=1.

The figure does not have a global minimum on the interval [0, 4].

Part (b) Step 1. Consider the function.

Consider the function $f\left(x\right)=\frac{\ln (x+3)}{x-3}$

the function is not defined at $x=3$, it is defined only in $x>-3$.

$f\text{'}\left(x\right)=\frac{1}{(x+3)(x-3)}-\frac{\ln (x+3)}{{(x-3)}^2}$

f'(x) is not defined at $x=-3,3$.

So there is no critical points and hence no global extrema.

As x tends to -3, the function approaches positive infinity, and as it tends to 3, the function approaches negative infinity.

So the function does not have a global extrema.

Part (c) Step 1. Find critical point.

Consider the function,

$$\begin{gathered} f\left(x\right)={(x-4)}^2 \\ f\text{'}\left(x\right)=2(x-4) \end{gathered}$$

The critical point is given by,

$$\begin{gathered} f\text{'}\left(x\right)=0 \\ x=4 \end{gathered}$$

which is a critical point.

Part (c) Step 2. Graph the function.

$f\text{'}\text{'}\left(x\right)=2>0$

Since $f\text{'}\left(4\right)=0,f\text{'}\text{'}\left(4\right)=2>0$, the function has a local minimum at x=4.

The values of function at endpoints is,

$$\begin{gathered} f\left(4\right)=0 \\ f\left(2\right)=4 \\ f\left(6\right)=4 \end{gathered}$$

Thus the global maximum occurs at $x=2,6$ and the global minimum occurs at $x=0$.

The graph of the function is: