Chapter 2.6, Problem 13E

To determine

To Prove: The value of $\Delta f$ and $\Delta g$ .

Explanation of Solution

Given:

The given functions are,

  $\begin{array}{l}f\left(x\right)=2x+3\\ g\left(x\right)=-3x+2\end{array}$

Calculation:

Consider the following function,

  $\begin{array}{l}f\left(x\right)=2x+3\\ g\left(x\right)=-3x+2\end{array}$

Since, the base points are $x_0=1.0$ and $\Delta x=0.1$ .

  $\begin{array}{c}\Delta f=f\left(x_0+\Delta x\right)-f\left(x_0\right)\\ =f\left(1.0+0.1\right)-f\left(1.0\right)\\ =\left(2\times 1.1\times 3\right)-\left(2\times 1.0+3\right)\\ =0.2\end{array}$

Since, the base points are $x_0=1.0$ and $\Delta x=0.1$ .

  $\begin{array}{c}\Delta g=g\left(x_0+\Delta x\right)-g\left(x_0\right)\\ =g\left(1.0+0.1\right)-g\left(1.0\right)\\ =\left(-3\times 1.1+2\right)-\left(-3\times 1.0+2\right)\\ =-0.3\end{array}$

Find the value of $\Delta \left(fg\right)$ by computing the value of $\Delta f\left(x_0+\Delta x\right)g\left(x_0+\Delta x\right)-f\left(x_0\right)g\left(x_0\right)$ as,

  $\begin{array}{c}\Delta \left(fg\right)=\Delta f\left(x_0+\Delta x\right)g\left(x_0+\Delta x\right)-f\left(x_0\right)g\left(x_0\right)\\ =f\left(1.0+0.1\right)g\left(1.0+0.1\right)-f\left(1.0\right)g\left(1.0\right)\\ =f\left(1.1\right)g\left(1.1\right)-f\left(1.0\right)g\left(1.0\right)\\ =\left[\left(2\times 1.1+3\right)\left(-3\times 1.1+2\right)\right]-\left[\left(2\times 1.0+3\right)\left(-3\times 1.0+2\right)\right]\end{array}$

Solve further as,

  $\begin{array}{c}\Delta \left(fg\right)=-6.76-\left(-5.0\right)\\ =-1.76\end{array}$

Find the value of $\Delta \left(fg\right)$ by computing the value of $g\left(x_0\right)\Delta f+f\left(x_0\right)\Delta g+\Delta f\Delta g$ as,

  $\begin{array}{c}\Delta \left(fg\right)=\left[g\left(1.0\right)\times 0.2\right]+\left[f\left(1.0\right)\times \left(-0.3\right)\right]+\left[0.2\times \left(-0.3\right)\right]\\ =\left[-3\left(1.0\right)\times 0.2\right]+\left[2\left(1.0\right)\times \left(-0.3\right)\right]+\left[0.2\times \left(-0.3\right)\right]\\ =-0.2-1.5-0.06\\ =-1.76\end{array}$

Therefore,

  $\Delta \left(fg\right)=g\left(x_0\right)\Delta f+f\left(x_0\right)\Delta g+\Delta f\Delta g$

Since, the base points are $x_0=1.0$ and $\Delta x=0.01$ .

  $\begin{array}{c}\Delta f=f\left(x_0+\Delta x\right)-f\left(x_0\right)\\ =f\left(1.0+0.01\right)-f\left(1.0\right)\\ =\left(2\times 1.01\times 3\right)-\left(2\times 1.0+3\right)\\ =0.02\end{array}$

Since, the base points are $x_0=1.0$ and $\Delta x=0.01$ .

  $\begin{array}{c}\Delta g=g\left(x_0+\Delta x\right)-g\left(x_0\right)\\ =g\left(1.0+0.01\right)-g\left(1.0\right)\\ =\left(-3\times 1.01+2\right)-\left(-3\times 1.0+2\right)\\ =-0.03\end{array}$

Find the value of $\Delta \left(fg\right)$ by computing the value of $\Delta f\left(x_0+\Delta x\right)g\left(x_0+\Delta x\right)-f\left(x_0\right)g\left(x_0\right)$ as,

  $\begin{array}{c}\Delta \left(fg\right)=\Delta f\left(x_0+\Delta x\right)g\left(x_0+\Delta x\right)-f\left(x_0\right)g\left(x_0\right)\\ =f\left(1.0+0.01\right)g\left(1.0+0.01\right)-f\left(1.0\right)g\left(1.0\right)\\ =f\left(1.01\right)g\left(1.01\right)-f\left(1.0\right)g\left(1.0\right)\\ =\left[\left(2\times 1.01+3\right)\left(-3\times 1.01+2\right)\right]-\left[\left(2\times 1.0+3\right)\left(-3\times 1.0+2\right)\right]\end{array}$

Solve further as,

  $\begin{array}{c}\Delta \left(fg\right)=-5.1706-\left(-5.0\right)\\ =-0.1706\end{array}$

Find the value of $\Delta \left(fg\right)$ by computing the value of $g\left(x_0\right)\Delta f+f\left(x_0\right)\Delta g+\Delta f\Delta g$ as,

  $\begin{array}{c}\Delta \left(fg\right)=\left[g\left(1.0\right)\times 0.02\right]+\left[f\left(1.0\right)\times \left(-0.03\right)\right]+\left[0.02\times \left(-0.03\right)\right]\\ =\left[-3\left(1.0\right)\times 0.02\right]+\left[2\left(1.0\right)\times \left(-0.03\right)\right]+\left[0.02\times \left(-0.03\right)\right]\\ =-0.2-1.5-0.06\\ =-1.706\end{array}$

Therefore,

  $\Delta \left(fg\right)=g\left(x_0\right)\Delta f+f\left(x_0\right)\Delta g+\Delta f\Delta g$

Hence, this shows that with $\Delta x=0.01$ the term $\Delta f\Delta g=-0.0006$ that is very small.