Chapter 4.5, Problem 28E

a.

To determine

To find: To find the true change in the given function.

Answer to Problem 28E

The true change in the given function $f(x)=x^3-x$ , $a=1$ and $\Delta x=0.1$ is $\Delta f=12.21$ .

Explanation of Solution

Given: The function $f(x)=x^3-x$ , $a=1$ , $\Delta x=0.1$ .

Formula used: $\Delta f=f(a+\Delta x)-f(a)$

Calculation:

The given function is $f(x)=x^3-x$ .

The true change can be found using the formula $\Delta f=f(a+\Delta x)-f(a)$.

The given conditions are $a=1$ , $\Delta x=0.1$ .

By substituting the above values in the formula,

  $\begin{array}{c}\Delta f=f(1+0.1)-f(1)\\ \Delta f=f(1.1)-f(1)\\ $ f(1.1)={(1.1)}^3-(1.1)$f(1.1)=13.31-1.1\\ f(1.1)=12.21\\ f(1)=1^3-1\\ f(1)=0\\ \Delta f=f(1.1)-f(0)\\ \Delta f=12.21-0\\ \Delta f=12.21\end{array}$

Conclusion:

The true change in the given function $f(x)=x^3-x$ , $a=1$ and $\Delta x=0.1$ is $\Delta f=12.21$ .

b.

To determine

To find: To find the estimated change in the given function.

Answer to Problem 28E

The estimated change in the given function $f(x)=x^3-x$ , $a=1$ and $\Delta x=0.1$ is $\Delta f=0.2$ .

Explanation of Solution

Given: The function $f(x)=x^3-x$ , $a=1$ , $\Delta x=0.1$ .

Formula used: $\Delta f=f\text{'}(a)\Delta x$

Calculation:

The given function is $f(x)=x^3-x$ .

The estimated change can be found using the formula $\Delta f=f\text{'}(a)\Delta x$.

The given conditions are $a=1$ , $\Delta x=0.1$ .

By substituting the above values in the formula,

  $\Delta f=f\text{'}(1)(0.1)$

Differentiating the function with respect to x,

  $f(x)=3x^2-1$

Substituting $x=0$ in the above equation,

  $\begin{array}{l}f\text{'}(1)=3{(}^1-1\\ f\text{'}(1)=2\end{array}$

Substituting in the formula,

  $\begin{array}{l}\Delta f=(2)(0.1)\\ \Delta f=0.2\end{array}$

Conclusion:

The estimated change in the given function $f(x)=x^3-x$ , $a=1$ and $\Delta x=0.1$ is $\Delta f=0.2$ .

c.

To determine

To find: To find the approximate error in the given function.

Answer to Problem 28E

The approximate error in the given function $f(x)=x^3-x$ , $a=1$ and $\Delta x=0.1$ is $\left|\Delta f-f\text{'}(a)\Delta x\right|=12.01$ .

Explanation of Solution

Given: The function $f(x)=x^3-x$ , $a=1$ , $\Delta x=0.1$ .

Formula used: $\left|\Delta f-f\text{'}(a)\Delta x\right|$

Calculation:

The given function is $f(x)=x^3-x$ .

The true change can be found using the formula $\left|\Delta f-f\text{'}(a)\Delta x\right|$.

The given conditions are $a=1$ , $\Delta x=0.1$ ,

By substituting the above values in the formula,

  $\begin{array}{c}\Delta f=f(1+0.1)-f(1)\\ \Delta f=f(1.1)-f(1)\\ $ f(1.1)={(1.1)}^3-(1.1)$f(1.1)=13.31-1.1\\ f(1.1)=12.21\\ f(1)=1^3-1\\ f(1)=0\\ \Delta f=f(1.1)-f(0)\\ \Delta f=12.21-0\\ \Delta f=12.21\end{array}$

  $\begin{array}{l}f\text{'}(a)\Delta x\\ f\text{'}(a)\Delta x=f\text{'}(0)(0.1)\\ f\text{'}(a)\Delta x=(2)(0.1)\\ f\text{'}(a)\Delta x=0.2\end{array}$

Approximate error will be

  $\begin{array}{l}\left|\Delta f-f\text{'}(a)\Delta x\right|=12.21-0.2\\ \left|\Delta f-f\text{'}(a)\Delta x\right|=12.01\end{array}$

Conclusion:

The approximate error in the given function $f(x)=x^3-x$ , $a=1$ and $\Delta x=0.1$ is $\left|\Delta f-f\text{'}(a)\Delta x\right|=0.01$ .