Chapter 2.6, Problem 4E

(a)

To determine

To find: The slope of the tangent line to the curve $y=x-x^3$ at the point $\left(1,0\right)$.

Answer to Problem 4E

The slope of the tangent line to the curve $y=x-x^3$ at the point $\left(1,0\right)$

(i) Using Definition 1 is $\underset{\_}{m=-2}$.

(ii) Using Equation 2 is $\underset{\_}{m=-2}$.

Explanation of Solution

Given:

The equation of the curve $y=x-x^3$.

The slope of the tangent line to the curve at the point $\left(1,0\right)$.

Formula used:

Definition 1: The slope of the tangent curve $y=f\left(x\right)$ at the point $P\left(a,f\left(a\right)\right)$, $m=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$.

Equation 2: The slope of the tangent line in definition 1 becomes, $m=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$.

Difference of square formula: $\left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)$.

Calculation:

Section (i)

Obtain the slope of the tangent line to the curve at the point (1, 0) using Definition 1.

Substitute $a=1$ and $f\left(a\right)=0$ in Definition 1,

$\begin{array}{c}m=\underset{x\to 1}{\lim }\frac{f\left(x\right)-0}{x-1}\\ =\underset{x\to 1}{\lim }\frac{x-x^3-0}{x-1}\\ =\underset{x\to 1}{\lim }\frac{x\left(1-x^2\right)}{x-1}\end{array}$

Apply the difference of squares formula,

$\begin{array}{c}m=\underset{x\to 1}{\lim }\frac{x\left(1-x\right)\left(1+x\right)}{x-1}\\ =\underset{x\to 1}{\lim }\frac{x\left(-\left(x-1\right)\right)\left(1+x\right)}{\left(x-1\right)}\\ =\underset{x\to 1}{\lim }\frac{-x\left(x-1\right)\left(1+x\right)}{\left(x-1\right)}\end{array}$

Since the limit x approaches 1 but not equal to 1, cancel the common term $\left(x-1\right)\left(\ne 0\right)$ from both the numerator and the denominator,

$\begin{array}{c}m=\underset{x\to 1}{\lim }\left(-x\left(x+1\right)\right)\\ =\underset{x\to 1}{\lim }\left(-x^2-x\right)\\ =-{\left(1\right)}^2-1\\ =-2\end{array}$

Thus, the slope of the tangent line to the parabola at the point (1, 0) by using Definition 1 is $\underset{\_}{m=-2}$.

Section (ii)

Obtain the slope of the tangent line to the parabola at the point (1, 0) by using Equation 2.

Substitute $a=1$ and $f\left(a\right)=0$ in Equation 2,

$\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{{\left(x-x^3\right)}_{x=1+h}-0}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(\left(1+h\right)-{\left(1+h\right)}^3\right)}{h}\end{array}$

Simplify the terms by expanding $\left(1+h^3\right)$ as shown below.

$\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{\left(\left(1+h\right)-\left(h^3+3h^2+3h+1\right)\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{1+h-h^3-3h^2-3h-1}{h}\\ =\underset{h\to 0}{\lim }\frac{-h^3-3h^2-2h}{h}\\ =\underset{h\to 0}{\lim }\frac{-h\left(h^2+3h+2\right)}{h}\end{array}$

Since the limit h approaches 0 but not equal to 0, cancel the common term $h\left(\ne 0\right)$ from both the numerator and the denominator,

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

Thus, the slope of the tangent line to the parabola at the point (1, 0) by using Equation 2 is $\underset{\_}{m=-2}$.

(b)

To determine

To find: The equation of the tangent line in part (a).

Answer to Problem 4E

The equation of the tangent line in part (a) is $\underset{\_}{y=-2x+2}$.

Explanation of Solution

Formula used:

The equation of the tangent line to the curve $y=f\left(x\right)$ at the point $\left(a,f\left(a\right)\right)$ is

$y-f\left(a\right)=f^{\prime }\left(a\right)\left(x-a\right)$ .

Calculation:

Obtain the equation of the tangent line in part (a).

Since the tangent line to the curve $y=f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ is the line through the point $\left(a,f\left(a\right)\right)$ whose slope is equal to the derivative of a, $f^{\prime }\left(a\right)=-2$.

Substitute $a=1$, $f\left(a\right)=0$ and the slope $f^{\prime }\left(a\right)=-2$ in the equation of tangent line,

$\begin{array}{c}\left(y-f\left(a\right)\right)=f^{\prime }\left(a\right)\left(x-a\right)\\ y-0=-2\left(x-1\right)\\ =-2x+2\\ \end{array}$

Isolate y, $y=-2x+2$.

Thus, the equation of the tangent line is $\underset{\_}{y=-2x+2}$.

(c)

To determine

To sketch: The graph of the tangent line and the curve. Check zoom in the graph toward the point (1, 0) until the curve and the tangent line is indistinguishable.

Explanation of Solution

The equation of the tangent line $y=-2x+2$

The equation of the given curve is $y=x-x^3$.

Use the online graphing calculator to draw the graph of the functions as shown below in Figure 1.

Graph:

Use the online graphing calculator to zoom toward the point (1, 0) as shown below in Figure 2.

From Figure 2, it is observed that when zoom in the graph toward the point (1, 0), the graph of the tangent line and the curve looks likes almost identical.

Hence, it is verified that the graph of the tangent line and the curve zooming toward the point (1, 0) until the tangent line are indistinguishable.