Chapter 11.4, Problem 49E

Solution

a.

The derivative of $f(x)=2x^2+3x+1$ .

  $f^{\prime }(x)=4x+3$

Given information:

The function $f(x)=2x^2+3x+1$ .

Calculation:

Find the derivative of the function $f(x)=2x^2+3x+1$ .

  $\begin{array}{c}{f}^{\prime }(x)=\frac{d}{dx}\left(2x^2+3x+1\right)\\ =4x+3\end{array}$

Therefore, the derivative of the function $f(x)=2x^2+3x+1$ is $f^{\prime }(x)=4x+3$ .

b.

The derivative of $g(x)=x^3+1$ .

  $g^{\prime }(x)=3x^2$

Given information:

The function $g(x)=x^3+1$ .

Calculation:

Find the derivative of the function $g(x)=x^3+1$ .

  $\begin{array}{c}{g}^{\prime }(x)=\frac{d}{dx}\left(x^3+1\right)\\ =3x^2+0\\ =3x^2\end{array}$

Therefore, the derivative of the function $g(x)=x^3+1$ is $g^{\prime }(x)=3x^2$ .

c.

The standard quotient $\frac{f(x+h)-f(x)}{h}$ and the symmetric quotient $\frac{f(x+h)-f(x-h)}{2h}$ using $x=2$ and $h=0.001$ .

  $\frac{f(x+h)-f(x)}{h}=11.002$ and $\frac{f(x+h)-f(x-h)}{2h}=11$ .

Given information:

The function $f(x)=2x^2+3x+1$ .

Calculation:

Find the standard quotient $\frac{f(x+h)-f(x)}{h}$ using $x=2$ and $h=0.001$ .

  $\begin{array}{c}\frac{f(x+h)-f(x)}{h}=\frac{f\left(2+0.001\right)-f\left(2\right)}{0.001}\\ =\frac{f\left(2.001\right)-f\left(2\right)}{0.001}\\ =\frac{\text{15}\text{.011002}-\text{15}}{0.001}\\ =\frac{\text{0}\text{.011002}}{0.001}\\ =11.002\end{array}$

Now, find the symmetric quotient $\frac{f(x+h)-f(x-h)}{2h}$ .

  $\begin{array}{c}\frac{f(x+h)-f(x-h)}{2h}=\frac{f\left(2+0.001\right)-f\left(2-0.001\right)}{2\times 0.001}\\ =\frac{\text{15}\text{.011002}-\text{14}\text{.989002}}{0.002}\\ =\frac{0.022}{0.002}\\ =11\end{array}$

d.

To compare the approximations to $f^{\prime }\left(2\right)$ in part (c) using $x=2$ and $h=0.001$ .

Symmetric quotient is the better approximation to $f^{\prime }\left(2\right)$ .

Given information:

The function $f(x)=2x^2+3x+1$ .

Calculation:

From part (c), $\frac{f(x+h)-f(x)}{h}=11.002$ and $\frac{f(x+h)-f(x-h)}{2h}=11$ using $x=2$ and $h=0.001$ .

Now, find $f^{\prime }\left(2\right)$ .

  $\begin{array}{c}{f}^{\prime }(2)=4\left(2\right)+3\\ =8+3\\ =11\end{array}$

Thus, it can be observed that symmetric quotient is the better approximation to $f^{\prime }\left(2\right)$ .

e.

The standard quotient and the symmetric quotient for $g(x)=x^3+1$ using $x=2$ and $h=0.001$ , can compare the result with $g^{\prime }(2)$ .

  $\frac{g(x+h)-g(x)}{h}=12.006001$ , $\frac{g(x+h)-g(x-h)}{2h}=12.000001$ and symmetric quotient is the better approximation to $g^{\prime }(2)$ .

Given information:

The function $g(x)=x^3+1$ .

Calculation:

Find the standard quotient $\frac{g(x+h)-g(x)}{h}$ using $x=2$ and $h=0.001$ .

  $\begin{array}{c}\frac{g(x+h)-g(x)}{h}=\frac{g\left(2+0.001\right)-g\left(2\right)}{0.001}\\ =\frac{g\left(2.001\right)-g\left(2\right)}{0.001}\\ =\frac{\text{9}\text{.012006001}-9}{0.001}\\ =\frac{\text{0}\text{.012006001}}{0.001}\\ =12.006001\end{array}$

Now, find the symmetric quotient $\frac{g(x+h)-g(x-h)}{2h}$ .

  $\begin{array}{c}\frac{g(x+h)-g(x-h)}{2h}=\frac{g\left(2+0.001\right)-g\left(2-0.001\right)}{2\times 0.001}\\ =\frac{\text{9}\text{.012006001}-\text{8}\text{.988005999}}{0.002}\\ =\frac{0.024000002}{0.002}\\ =12.000001\end{array}$

Now, find $g^{\prime }(2)$ .

  $\begin{array}{c}{g}^{\prime }(2)=3{\left(2\right)}^2\\ =3\times 4\\ =12\end{array}$

Thus, it can be observed that symmetric quotient is the better approximation to $g^{\prime }(2)$ .