Q. 16 - 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.

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

$$\begin{gathered} \left(a\right) (-5,2] \\ \left(b\right) [-5,2) \\ \left(c\right) [1,10] \\ \left(d\right) [0,20] \end{gathered}$$

Step-by-Step Solution

Answer

(a) The global maximum of the function $f\left(x\right)=\frac{4}{\sqrt{x^2+1}}+3$ is $x=0,2$ and at the values $f\left(0\right)=7, f\left(2\right)=4.78$

(b) The global maximum of the function is $x=-5,0$ and at the values $f(-5)=3.78, f\left(0\right)=7$.

(c) The global maximum of the function is $x=1,10$ and at the values $f\left(1\right)=5.8, f\left(10\right)=3.39$. 

(d) The global maximum of the function is $x=0,20$ and at the values $f\left(0\right)=7, f\left(20\right)=3.2$. 

Part (a) Step 1. Given Information.

The function: 

$f\left(x\right)=\frac{4}{\sqrt{x^2+1}}+3$

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

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

So,

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

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

The critical points can tested as:

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{8x^2-4}{{(x^2+1)}^{{\displaystyle \frac{5}{2}}}} \\ f\text{'}\text{'}\left(0\right)=-4<0 \end{gathered}$$

So the function has a local maximum at $x=0$

The height of the local extrema is,

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

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

Find the global extrema in the interval $(-5,2]$

$$\begin{gathered} \underset{x\to -5^{+}}{lim}f\left(x\right)=\underset{x\to -5^{+}}{lim}(\frac{4}{\sqrt{x^2+1}}+3) \\ =3.78 \\ f\left(2\right)=\frac{4}{\sqrt{2^2+1}}+3 \\ =4.78 \end{gathered}$$

The global maximum is at $x=0,2 with f\left(0\right)=7, f\left(2\right)=4.78$.

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 $[-5,2)$.

$$\begin{gathered} f(-5)=\frac{4}{\sqrt{{(-5)}^2+1}}+3 \\ =3.78 \\ \underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}(\frac{4}{\sqrt{x^2+1}}+3) \\ =4.78 \end{gathered}$$

The global maximum is at $x=-5,0 with f(-5)=3.78, f\left(0\right)=7$.

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 $[1,10]$.

$$\begin{gathered} f\left(1\right)=\frac{4}{\sqrt{{\left(1\right)}^2+1}}+3 \\ =5.8 \\ f\left(10\right)=\frac{4}{\sqrt{{\left(10\right)}^2+1}}+3 \\ =3.39 \end{gathered}$$

The global maximum is at  $x=1,10 with f\left(1\right)=5.8, f\left(10\right)=3.39$.

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 $[0,20]$.

$$\begin{gathered} f\left(0\right)=\frac{4}{\sqrt{{\left(0\right)}^2+1}}+3 \\ =7 \\ f\left(20\right)=\frac{4}{\sqrt{{\left(20\right)}^2+1}}+3 \\ =3.2 \end{gathered}$$

The global maximum is $x=0,20 with f\left(0\right)=7, f\left(20\right)=3.2$.

Part (d) Step 2. Graph the function.

The graph of the function is: