Q. 19 - Calculus

Question

Find the locations and values of any global extrema of each function f in Exercises 11–20 on each of the four given intervals. Do all work by hand by considering local extrema and endpoint behavior. Afterwards, check your answers with graphs.

$f\left(x\right)={(x^2-4x+3)}^{-\frac{1}{2}}$ on the intervals,

$$\begin{gathered} \left(a\right) [0,4] \\ \left(b\right) [0,10] \\ \left(c\right) [0,3.5] \\ \left(d\right) (3,\infty ) \end{gathered}$$

Step-by-Step Solution

Answer

(a) There is no global maximum of the function $f\left(x\right)={(x^2-4x+3)}^{-\frac{1}{2}}$ and the global minimum at $x=0,4$ and at the values $f\left(0\right)=f\left(4\right)=\frac{1}{\sqrt{3}}$.

(b) There is no global maximum and the global minimum at $x=10$ and at the values $f\left(10\right)=\frac{1}{3\sqrt{7}}$.

(c) There is no global maximum and the global minimum at $x=0$ and at the values $f\left(0\right)=\frac{1}{\sqrt{3}}$.

(d) There is no global maximum and the global minimum.

Part (a) Step 1. Given Information.

The function: 

$$\begin{gathered} f\left(x\right)={(x^2-4x+3)}^{-\frac{1}{2}} \\ [0,4] \end{gathered}$$

Part (a) Step 2. Find the critical points.

The critical points are given by,
$$\begin{gathered} f\left(x\right)={(x^2-4x+3)}^{-\frac{1}{2}} \\ f\text{'}\left(x\right)=\frac{2-x}{{(x^2-4x+3)}^{{\displaystyle \frac{3}{2}}}} \end{gathered}$$

So,

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

Part (a) Step 3. Test the critical points.

The critical points can tested as: 

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{\sqrt{x^2-4x+3}(2x^2-8x+9)}{{(x-3)}^3{(x-1)}^3} \\ f\text{'}\text{'}\left(2\right)=-1<0 \end{gathered}$$

So the function has a local maximum at $x=2$

The height of the local extrema is, 

$$\begin{gathered} f\left(2\right)={({\left(2\right)}^2-4(2)+3)}^{-\frac{1}{2}} \\ =Undefined \end{gathered}$$

Part (a) Step 4. Check the height at endpoint values.

Find the global extrema in the interval $[0,4]$.

$$\begin{gathered} f\left(0\right)={({\left(0\right)}^2-4(0)+3)}^{-\frac{1}{2}} \\ =\frac{1}{\sqrt{3}} \\ f\left(4\right)={({\left(4\right)}^2-4(4)+3)}^{-\frac{1}{2}} \\ =\frac{1}{\sqrt{3}} \end{gathered}$$

There is no global maximum and the global minimum is at $x=0,4 with f\left(0\right)=f\left(4\right)=\frac{1}{\sqrt{3}}$.

Part (a) Step 5. Sketch the graph.

The graph of the function is: 

Part (b) Step 1. Check the height at endpoint values.

Find the global extrema in the interval $[0,10]$.

$$\begin{gathered} f\left(0\right)={({\left(0\right)}^2-4(0)+3)}^{-\frac{1}{2}} \\ =\frac{1}{\sqrt{3}} \\ f\left(10\right)={({\left(10\right)}^2-4(10)+3)}^{-\frac{1}{2}} \\ =\frac{1}{3\sqrt{7}} \end{gathered}$$

There is no global maximum and the global minimum is at $x=10 with f\left(10\right)=\frac{1}{3\sqrt{7}}$.

Part (b) Step 2. Graph the function.

The graph of the function is: 

Part (c) Step 1. Check the height at endpoint values.

Find the global extrema in the interval $[0,3.5]$.

$$\begin{gathered} f\left(0\right)={({\left(0\right)}^2-4(0)+3)}^{-\frac{1}{2}} \\ =\frac{1}{\sqrt{3}} \\ f(3.5)={({(3.5)}^2-4(3.5)+3)}^{-\frac{1}{2}} \\ =0.89 \end{gathered}$$

There is no global maximum and the global minimum is at $x=0 with f\left(0\right)=\frac{1}{\sqrt{3}}$. 

Part (c) Step 2. Graph the function.

The graph of the function is: 

Part (d) Step 1. Check the height at endpoint values.

Find the global extrema in the interval $(3,\infty )$.

$$\begin{gathered} \underset{x\to 3^{-}}{lim}f\left(3\right)={\underset{x\to 3^{-}}{lim}({\left(3\right)}^2-4(3)+3)}^{-\frac{1}{2}} \\ =0 \\ \underset{x\to {\infty }^{+}}{lim}=\underset{x\to {\infty }^{+}}{lim}{({(\infty )}^2-4(\infty )+3)}^{-\frac{1}{2}} \\ =\frac{1}{\sqrt{3}} \end{gathered}$$

There is no global maximum and the global minimum.

Part (d) Step 2. Graph the function.

The graph of the function is: