Chapter 2.7, Problem 18E

(a)

To determine

To estimate: The value of $f^{\prime }\left(0\right)$, $f^{\prime }\left(\frac{1}{2}\right)$, $f^{\prime }\left(1\right)$, $f^{\prime }\left(2\right)$ and $f^{\prime }\left(3\right)$ by using a graphing device to zoom in on the graph of f.

Answer to Problem 18E

The value of $f^{\prime }\left(0\right)$, $f^{\prime }\left(\frac{1}{2}\right)$, $f^{\prime }\left(1\right)$, $f^{\prime }\left(2\right)$ and $f^{\prime }\left(3\right)$ are 0, 0.75, 3, 12 and 27.

Explanation of Solution

Given:

The function is $f\left(x\right)=x^3$.

Estimation:

Obtain $f^{\prime }\left(x\right)$ at the point 0.

Use the online graphing calculator to zoom toward the point $\left(0,0\right)$ as shown below in Figure 1.

The calculation of $f^{\prime }\left(0\right)$ is as follows,

From Figure 1, the tangent to the curve at $x=0$ is the x-axis.

So, $f^{\prime }\left(0\right)=0$.

Thus, $f^{\prime }\left(0\right)=0$.

Obtain $f^{\prime }\left(x\right)$ at the point $\frac{1}{2}$.

Use the online graphing calculator to zoom toward the point $\left(\frac{1}{2},\frac{1}{8}\right)$ as shown below in Figure 2.

The calculation of $f^{\prime }\left(\frac{1}{2}\right)$ is as follows,

From Figure 2,the slope of the tangent is as follows,

$\begin{array}{c}m=\frac{0.05-0.2}{0.4-0.6}\\ =\frac{-0.15}{-0.2}\\ =\frac{15}{20}\\ =0.75\end{array}$

Thus, $f^{\prime }\left(\frac{1}{2}\right)=0.75$.

Obtain $f^{\prime }\left(x\right)$ at the point 1.

Use the online graphing calculator to zoom toward the point $\left(1,1\right)$ as shown below in Figure 3.

The calculation of $f^{\prime }\left(1\right)$ is as follows.

From Figure 3, the slope of tangent is,

$\begin{array}{c}m=\frac{0.4-1.3}{0.8-1.1}\\ =\frac{0.9}{0.3}\\ =3\end{array}$

Thus, $f^{\prime }\left(1\right)=3$.

Obtain $f^{\prime }\left(x\right)$ at the point 2.

Use the online graphing calculator to zoom toward the point $\left(2,8\right)$ as shown below in Figure 4.

The calculation of $f^{\prime }\left(2\right)$ is as follows.

From Figure 4,the slope of tangent is

$\begin{array}{c}m=\frac{4.4-12.8}{1.7-2.4}\\ =\frac{-8.4}{-0.7}\\ =12\end{array}$

Thus, $f^{\prime }\left(2\right)=12$.

Obtain $f^{\prime }\left(x\right)$ at the point 3.

Use the online graphing calculator to zoom toward the point $\left(3,27\right)$ as shown below in Figure 4.

The calculation of $f^{\prime }\left(3\right)$ is as follows,

From Figure 5, the slope of tangent is

$\begin{array}{c}m=\frac{43.2-16.2}{3.6-2.6}\\ =\frac{27}{1}\\ =27\end{array}$

Thus, $f^{\prime }\left(3\right)=27$.

(b)

To determine

To deduce: The values of $f^{\prime }\left(-\frac{1}{2}\right),f^{\prime }\left(-1\right),f^{\prime }\left(-2\right)$ and $f^{\prime }\left(-3\right)$ using symmetry.

Answer to Problem 18E

The values of $f^{\prime }\left(-\frac{1}{2}\right),f^{\prime }\left(-1\right),f^{\prime }\left(-2\right)$ and $f^{\prime }\left(-3\right)$ are 0.75, 3, 12 and 27 respectively.

Explanation of Solution

Result Used:

For any odd and even function,

$f\left(-x\right)=-f\left(x\right)$ and $f\left(-x\right)=f\left(x\right)$ holds respectively.

Calculation:

The function $f\left(x\right)=x^3$ is odd function.

Thus, the function $f^{\prime }\left(x\right)$ is an even function.

From part (a),

$f^{\prime }\left(\frac{1}{2}\right)=0.75,f^{\prime }\left(1\right)=1,f^{\prime }\left(2\right)=12$ and $f^{\prime }\left(3\right)=27$.

Using the symmetry,

$\begin{array}{c}{f}^{\prime }\left(-\frac{1}{2}\right)=f^{\prime }\left(\frac{1}{2}\right)\\ =0.75\end{array}$

Thus, $f^{\prime }\left(-\frac{1}{2}\right)=0.75$.

Using the symmetry,

$\begin{array}{c}{f}^{\prime }\left(-1\right)=f^{\prime }\left(1\right)\\ =3\end{array}$

Thus, $f^{\prime }\left(-1\right)=3$.

Using the symmetry,

$\begin{array}{c}{f}^{\prime }\left(-2\right)=f^{\prime }\left(2\right)\\ =12\end{array}$

Thus, $f^{\prime }\left(-2\right)=12$.

Using the symmetry,

$\begin{array}{c}{f}^{\prime }\left(-3\right)=f^{\prime }\left(3\right)\\ =27\end{array}$

Thus, $f^{\prime }\left(-3\right)=27$.

(c)

To determine

To sketch: The graph of $f^{\prime }$ using parts a and b.

Explanation of Solution

Graph:

Use the information from part a and b and sketch the graph of $f^{\prime }$ as shown in Figure 6.

From Figure 6, it is observed that the function $f^{\prime }$ is an even function.

(d)

To determine

To guess: The formula of $f^{\prime }$ by using part a and part b.

Answer to Problem 18E

The formula of $f^{\prime }$ is $f^{\prime }\left(x\right)=3x^2$

Explanation of Solution

From part a and part b,

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

$\begin{array}{c}{f}^{\prime }\left(-1\right)=3\\ =3{\left(-1\right)}^2\end{array}$

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

$\begin{array}{c}{f}^{\prime }\left(-3\right)=27\\ =3\left(9\right)\\ =3{\left(-3\right)}^2\end{array}$

It is observed from the above calculations that the derivative is thrice the square of the input value.

Thus, the formula for the function is, $f^{\prime }\left(x\right)=3x^2$.

(e)

To determine

To prove: The guess in part d by using the definition of derivative.

Explanation of Solution

Definition used:

The derivation of a function is given by the formula,

$f^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Proof:

Consider the function, $f\left(x\right)=x^3$.

Use the definition of derivative to obtain the derivative of $f\left(x\right)=x^3$.

$\begin{array}{c}{f}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{{\left(x+h\right)}^3-{\left(x\right)}^3}{h}\\ =\underset{h\to 0}{\lim }\frac{x^3+h^3+3x^{2}h+3x{h}^2-x^3}{h}\end{array}$

Simplify the terms in numerator,

$f^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{h^3+3x^{2}h+3x{h}^2}{h}$

Since the limit h approaches zero but not equal to zero, cancel the common term h from both the numerator and the denominator,

$\begin{array}{c}{f}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\left(h^2+3x^2+3xh\right)\\ =\underset{h\to 0}{\lim }\left(h^2\right)+3x^2\underset{h\to 0}{\lim }\left(1\right)+3x\underset{h\to 0}{\lim }\left(h\right)\\ =0+3x^2+3x\left(0\right)\\ =3x^2\end{array}$

So, $f^{\prime }\left(x\right)=3x^2$.

Thus, the guess in part d is correct.