Chapter 2.7, Problem 17E

(a)

To determine

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

Answer to Problem 17E

The value of $f^{\prime }\left(0\right)$, $f^{\prime }\left(\frac{1}{2}\right)$, $f^{\prime }\left(1\right)$ and $f^{\prime }\left(2\right)$ are 0, 1, 2 and 4 respectively.

Explanation of Solution

Given:

The function is $f\left(x\right)=x^2$.

Estimation:

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

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.

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

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

Use the online graphing calculator to zoom toward the point $\left(\frac{1}{2},\frac{1}{4}\right)$ 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 line to the curve is computed as follows.

$\begin{array}{c}m=\frac{0.3536-0.164}{0.6036-0.414}\\ \approx 1\end{array}$

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

Obtain the value of $f^{\prime }\left(x\right)$ 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 the tangent line to the curve is computed as follows.

$\begin{array}{c}m=\frac{1.5-0.6}{1.25-0.8}\\ =\frac{0.9}{0.45}\\ =2\end{array}$

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

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

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

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

From Figure 4, the slope of the tangent line is computed as follows.

$\begin{array}{c}m=\frac{2-5}{1.5-2.25}\\ =\frac{-3}{-0.75}\\ =\frac{-3}{-\frac{3}{4}}\\ =4\end{array}$

Thus, $f^{\prime }\left(2\right)=4$.

(b)

To determine

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

Answer to Problem 17E

The values of $f^{\prime }\left(-\frac{1}{2}\right),f^{\prime }\left(-1\right)$ and $f^{\prime }\left(-2\right)$ are $-1,-2\text{ and }-4$ respectively.

Explanation of Solution

Result Used:

For an odd function f, $f\left(-x\right)=-f\left(x\right)$.

For an even function f, $f\left(-x\right)=f\left(x\right)$.

Calculation:

The function $f\left(x\right)=x^2$ is an even function.

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

From part (a), $f^{\prime }\left(\frac{1}{2}\right)=1,f^{\prime }\left(1\right)=2$ and $f^{\prime }\left(2\right)=4$.

Using the symmetry,

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

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

Using the symmetry,

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

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

Using the symmetry,

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

Thus, $f^{\prime }\left(-2\right)=-4$.

Therefore, the values of $f^{\prime }\left(-\frac{1}{2}\right),f^{\prime }\left(-1\right)$ and $f^{\prime }\left(-2\right)$ are $-1,-2\text{ and }-4$ respectively.

(c)

To determine

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

Answer to Problem 17E

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

Explanation of Solution

From part (a) and part (b),

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

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

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

From the above calculations, it is observed that the derivative of the function is twice the input value.

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

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

(d)

To determine

To prove: The answer in part (c) is correct 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 $f\left(x\right)=x^2$.

Use the definition of derivative to obtain the derivative of $f\left(x\right)=x^2$ as follows.

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

Simplify the terms in numerator,

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

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

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

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

Thus, the guess in part (c) is correct.