Chapter 10.3, Problem 21E

To determine

The value of x when x < sinx.

Answer to Problem 21E

The inequality $x<\sin x\text{is true for }x<0$ , as may see by graphing $y=\sin x-x$

  $\left[-{10}^{-3},{10}^{-3}\right]\text{By}\left[-2\times {10}^{-10},2\times {10}^{-10}\right]$

Explanation of Solution

Given information:

The given function is

  $f\left(x\right)=\sin x$

Then,

  $P_1\left(x\right)=x$

Formula used:

Remainder Estimation Theorem is used.

Calculation:

Using Remainder Estimation Theorem for $n=3$ (use $n=3$ instead $n=2$ in this way better range of x)

Since,

  $P_2\left(x\right)=x$

Now,

  $f^{\left(3\right)}\left(x\right)=-\cos x$

Therefore,

  $\left|f^{\left(3\right)}\left(x\right)\right|=\left|-\cos x\right|\le 1$ for all real x

Use $M=r=1$ , which gives $\left|R_2\left(x\right)\right|\le \frac{{\left|x\right|}^3}{3!}$ , where $\left|x\right|<{10}^{-3}$

Therefore,

Maximum error is $\begin{array}{l}\frac{{\left|x\right|}^3}{3!}=\frac{{\left({10}^{-3}\right)}^3}{3!}=1.67\times {10}^{-10}\\ \text{error}=1.67\times {10}^{-10}\end{array}$

Alternate method: Using graphing technique, find that when

  $\left|x\right|<{10}^3.\left|\text{error}\right|=\left|\sin x-x\right|$

  $<\left|\sin {10}^{-3}-{10}^{-3}\right|=1.67\times {10}^{-10}$

The inequality $x<\sin x\text{is true for }x<0$ , as may see by graphing $y=\sin x-x$

  $\left[-{10}^{-3},{10}^{-3}\right]\text{By}\left[-2\times {10}^{-10},2\times {10}^{-10}\right]$

Conclusion:

The inequality $x<\sin x\text{is true for }x<0$ , as may see by graphing $y=\sin x-x$

  $\left[-{10}^{-3},{10}^{-3}\right]\text{By}\left[-2\times {10}^{-10},2\times {10}^{-10}\right]$