Chapter 2.6, Problem 14E

To determine

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

Explanation of Solution

Given:

The given functions are,

  $\begin{array}{l}f\left(x\right)=x^2+2\\ g\left(x\right)=3x^2-1\end{array}$

Calculation:

Consider the following function,

  $\begin{array}{l}f\left(x\right)=x^2+2\\ g\left(x\right)=3x^2-1\end{array}$

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

  $\begin{array}{c}\Delta f=f\left(x_0+\Delta x\right)-f\left(x_0\right)\\ =f\left(1.0+0.1\right)-f\left(1.0\right)\\ =\left[{\left(1.1\right)}^2+2\right]-\left[{\left(1.0\right)}^2+2\right]\\ =0.21\end{array}$

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

  $\begin{array}{c}\Delta g=g\left(x_0+\Delta x\right)-g\left(x_0\right)\\ =g\left(1.0+0.1\right)-g\left(1.0\right)\\ =\left(3\times {\left(1.1\right)}^2+2\right)-\left(3\times {\left(1.0\right)}^2-1\right)\\ =0.63\end{array}$

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,

  $\begin{array}{c}\Delta \left(fg\right)=\Delta f\left(x_0+\Delta x\right)g\left(x_0+\Delta x\right)-f\left(x_0\right)g\left(x_0\right)\\ =f\left(1.0+0.1\right)g\left(1.0+0.1\right)-f\left(1.0\right)g\left(1.0\right)\\ =f\left(1.1\right)g\left(1.1\right)-f\left(1.0\right)g\left(1.0\right)\\ =\left[\left({\left(1.1\right)}^2+2\right)\left(3\times {\left(1.1\right)}^2-1\right)\right]-\left[\left({\left(1.1\right)}^2+2\right)\left(3\times {\left(1.0\right)}^2-1\right)\right]\end{array}$

Solve further as,

  $\begin{array}{c}\Delta \left(fg\right)=\left(3.21\right)\left(2.63\right)-\left(3.0\right)\left(2.0\right)\\ =2.4423\end{array}$

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,

  $\begin{array}{c}\Delta \left(fg\right)=\left[g\left(1.0\right)\times 0.21\right]+\left[f\left(1.0\right)\times \left(0.63\right)\right]+\left[0.21\times \left(0.63\right)\right]\\ =\left[\left(3{\left(1.0\right)}^2-1\right)0.21\right]+\left[\left({\left(1.0\right)}^2\times 2\right)0.63\right]+\left[0.21\times \left(0.63\right)\right]\\ =0.42+1.86+0.1323\\ =2.4423\end{array}$

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 $\Delta x=0.01$ .

  $\begin{array}{c}\Delta f=f\left(x_0+\Delta x\right)-f\left(x_0\right)\\ =f\left(1.0+0.01\right)-f\left(1.0\right)\\ =\left[{\left(1.01\right)}^2+2\right]-\left[{\left(1.0\right)}^2+2\right]\\ =0.0201\end{array}$

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

  $\begin{array}{c}\Delta g=g\left(x_0+\Delta x\right)-g\left(x_0\right)\\ =g\left(1.0+0.01\right)-g\left(1.0\right)\\ =\left(3\times {\left(1.01\right)}^2+2\right)-\left(3\times {\left(1.0\right)}^2-1\right)\\ =0.603\end{array}$

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,

  $\begin{array}{c}\Delta \left(fg\right)=\Delta f\left(x_0+\Delta x\right)g\left(x_0+\Delta x\right)-f\left(x_0\right)g\left(x_0\right)\\ =f\left(1.0+0.01\right)g\left(1.0+0.01\right)-f\left(1.0\right)g\left(1.0\right)\\ =f\left(1.01\right)g\left(1.01\right)-f\left(1.0\right)g\left(1.0\right)\\ =\left[\left({\left(1.01\right)}^2+2\right)\left(3\times {\left(1.01\right)}^2-1\right)\right]-\left[\left({\left(1.0\right)}^2+2\right)\left(3\times {\left(1.0\right)}^2-1\right)\right]\end{array}$

Solve further as,

  $\begin{array}{c}\Delta \left(fg\right)=\left(3.0201\right)\left(2.0603\right)-\left(3.0\right)\left(2.0\right)\\ =0.22231203\end{array}$

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,

  $\begin{array}{c}\Delta \left(fg\right)=\left[g\left(1.0\right)\times 0.0201\right]+\left[f\left(1.0\right)\times \left(0.0603\right)\right]+\left[0.0201\times \left(0.0603\right)\right]\\ =\left[\left(3{\left(1.0\right)}^2-1\right)0.0201\right]+\left[\left({\left(1.0\right)}^2+2\right)0.0603\right]+\left[0.0201\times \left(0.0603\right)\right]\\ =0.07402+0.1809+0.00121203\\ =0.22231203\end{array}$

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 $\Delta f\Delta g=0.00121203$ that is very small.