Chapter 10.2, Problem 35E

(a)

To determine

To graph $f\left(x\right)=\frac{4x}{x^2+1}$ together with its linearization at $x=0$ and $x=\sqrt{3}$

Explanation of Solution

Given information: The given functionis $f\left(x\right)=\frac{4x}{x^2+1}$ and its inflection points are

  $x=0$ and $x=\sqrt{3}$

Formula Used:

Linearization of a function f

• The Linearization of a differentiable functionf at a point $x=a$ is the linear function

  $L\left(x\right)=f\left(a\right)+f\text{'}\left(a\right)\left(x-a\right)$

• The graph of linearizationof fis the tangent line to the graph of f at the point $\left(a,f\left(a\right)\right)$ .

Calculation:

  $f\left(x\right)=\frac{4x}{x^2+1}$

  $f\text{'}\left(x\right)$ $=\frac{\left(x^2+1\right)\frac{d\left(4x\right)}{dx}-\left(4x\right)\frac{d\left(x^2+1\right)}{dx}}{{\left(x^2+1\right)}^2}$

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

  $=\frac{4x^2+4-8x^2}{{\left(x^2+1\right)}^2}$

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

For, $x=0$

  $f\left(0\right)=0$

  $f\text{'}\left(0\right)=4$

?The Linearization of fat a point $x=0$ is:

  $L\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)\left(x-0\right)$

  $=0+4\left(x-0\right)$

  $=4x$

For, $x=\sqrt{3}$

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

  $=\frac{4\sqrt{3}}{4}=\sqrt{3}$

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

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

  $=\frac{-8}{4^2}=\frac{-1}{2}$

?The Linearization of fat the point $x=\sqrt{3}$ is:

  $L_1\left(x\right)=f\left(\sqrt{3}\right)+f\text{'}\left(\sqrt{3}\right)\left(x-\sqrt{3}\right)$

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

  $=\frac{2\sqrt{3}-x+\sqrt{3}}{2}$

  $=\frac{3\sqrt{3}-x}{2}$

Graph:

  

	$4x$$\frac{4x}{x^2+1}$
	$4x$
	$\frac{3\sqrt{3}-x}{2}$

Interpretation:

From the graph, the linearization of $f\left(x\right)$ fits particularly well at the inflection points $x=0$ and $x=\sqrt{3}$

Also,linearization of fat a point $x=0$ and $x=\sqrt{3}$ are the tangents lines to the graph of f at

  $x=0$ and $x=\sqrt{3}$

(b)

To determine

To prove the linearization of f at $x=a$ is also the second order Taylor polynomial of f at $x=a$

Explanation of Solution

Given Information: $f\left(x\right)$ is a twice differentiable function and graph of $f\left(x\right)$ has inflection point at $x=a$

Formula used:

The Taylor series expansion of $f\left(x\right)$ around the point $x=a$ is given by,

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

Proof:

Slope of $f\left(x\right)$ at $x=a$ is $f\text{'}\left(a\right)$ .

Linearization of $f\left(x\right)$ at a is:

  $f\text{'}\left(a\right)=\frac{f\left(x\right)-f\left(a\right)}{x-a}$

  $f\left(x\right)-f\left(a\right)=xf\text{'}\left(a\right)-af\text{'}\left(a\right)$

  $f\left(x\right)=f\left(a\right)+f\text{'}\left(a\right)\left(x-a\right)$

Which is the Taylor Expansion of order 2 at $x=a$