Math 242: Calculus I

1.4 The Problem with Tangent Lines

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1.4.1 More About Sets on the Real Line

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Definition

1. An open interval of the real line is an interval that does not contain either of its end points.

2. A closed interval of the real line is an interval that contains both of its end points.

3. A half-open interval of the real line is an interval that contains one end point, but not the other.

Example 1

Open intervals are denoted using strict inequalities, round brackets, or open dot $\circ$ at the end of the graph of an interval.

If $a$ and $b$ are real numbers and $a\le b$, then the following are open intervals

$$ \begin{align*} (a,b) &= \left\{\,x\in\mathbb{R}\,:\,a < x < b\,\right\} \\ (-\infty,b) &= \left\{\,x\in\mathbb{R}\,|\,x < b\,\right\} \\ (a,+\infty) &= \left\{\,x\in\mathbb{R}\,:\,x > a\,\right\} \\ (-\infty,+\infty) &= \mathbb{R} \\ (a,a) &= \emptyset = \left\{\ \right\} \end{align*} $$

Example 2

The following are closed intervals

$$ \begin{align*} [a,b] &= \left\{\,x\in\mathbb{R}\,|\,a\le x\le b\,\right\} \\ [a,a] &= \left\{\,a\,\right\} \end{align*} $$

Example 3

The following are half-open intervals

$$ \begin{align*} (a,b] &= \left\{\,x\in\mathbb{R}\,:\,a < x\le b\,\right\} \\ [a,b) &= \left\{\,x\in\mathbb{R}\,|\,a\le x < b\,\right\} \\ [a,+\infty) &= \left\{\,x\in\mathbb{R}\,:\,a\le x\,\right\} \\ (-\infty,b] &= \left\{\,x\in\mathbb{R}\,|\,x\le b\,\right\} \end{align*} $$

Definition

A neighborhood of a point on the real line is an open interval containing that point and that lies in the set containing the point.

Example 4

• A neighborhood of a point in the domain of a function is a small open interval in the domain that contains the point.

• A neighborhood of a point in the codomain of a function is a small open interval in the codomain that contains the point.

Example 5

Consider the function $y=x^2$. The (largest) domain of this function is the real line $\mathbb{R}$. Any open interval contain zero is a neighborhood of $0$ because the domain is so large!

Figure 1

$$ (-3,1),\ (-1,3),\ (-2,2),\ \left(-\frac{1}{3},\frac{1}{27}\right) $$

These intervals are all neighborhoods of zero in the domain $\mathbb{R}$.

Example 6

Consider the functon $g(x)=\sqrt{x}$. The domain of $g$ is $[0,\infty)$. There are no neighborhoods of $0$ in this domain! However, the point $\frac{1}{8}$ has infinitely many small neighborhoods in the domain of $g$.

Figure 2

$$ \left(\frac{1}{16},\frac{3}{16}\right) $$

This is a neighborhood of the point $\frac{1}{8}$ in the domain of $g$.

$$ \left(\frac{1}{4},\frac{2}{5}\right) $$

This is a neighborhood of the point $\frac{1}{2\sqrt{2}}$ in the codomain of $g$.

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1.4.2 Secant and Tangent Lines

Definition

A secant line is a line that intersects a curve more than once.

Secant lines occur frequently. In differential calculus we are looking for a more specific line, a tangent line.

Definition

A tangent line is a line that intersects a curve exactly once in a neighborhood of the point in the domain.

Figure 3

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1.4.3 Finding the Equation of the Tangent Line

First let's find the equation of a secant line to a curve.

Example 7

Figure 4

In order to compute the equation of the line we need a point and the slope. The slope is computed with two known points on the secant line. Naturally we pick the ones that intersect the curve.

$$ (x_1,y_1) = \left(-\frac{1}{2}, \frac{1}{4}\right),\ \text{and}\ (x_2,y_2) = \left( 1, 1 \right) $$

Using the formula for slope

$$ m = \frac{y_2-y_1}{x_2-x_1} = \frac{1-\frac{1}{4}}{1-\left(-\frac{1}{2}\right)} = \frac{\frac{3}{4}}{\frac{3}{2}} = \frac{3}{4}\cdot\frac{2}{3} = \frac{1}{2} $$

Utilizing the point-slope formula

$$ \begin{align*} y - y_2 &= m(x - x_2) \\ y - 1 &= \frac{1}{2}\left(x - 1\right) = \frac{x-1}{2} \\ y &= \frac{x-1}{2} + 1 = \frac{x+1}{2} \end{align*} $$

Example 8

Figure 5

Houston! We have a problem!

Sure there are infinitely many other points on the curve; and there are infinitely many other points on the tangent line. However we have no way of determining them because there are no other points on both.

This is the Tangent Problem

We will compute the slopes of several secant lines that are close to the tangent line. The point on the curve and the tangent line $P = (1, 1) = (1,f(1)) = (1,1^2)$.

Figure 6

Now we simply need another point $Q = (x,y) = (x,f(x)) = (x,x^2)$ so the $x\neq 1$. We already have the equation of the secant line(s) for any $x\neq 1$,

$$ m = \frac{y_2-y_1}{x_2-x_1} = \frac{x^2 - 1^2}{x-1} = \frac{(x+1)(x-1)}{x-1} = x+1 \\ $$

This yields an equation for the slope of the secant line for any $x$ in a neighborhood of $x=1$. Let's compute the value of the slope of the secant line for several values for $x$.

x	m		x	m
0.5	1.5		1.5	2.5
0.75	1.75		1.25	2.25
0.90	1.90		1.1	2.1
0.95	1.95		1.05	2.05
0.99	1.99		1.01	2.01
0.999	1.999		1.001	2.001

Table 1

As the points in the domain get closer and closer to $1$, then the points on the curve $(x,x^2)$ gets closer to $(1,1)$. Likewise the secant line get closer and closer to the tangent line.

Inspecting Table 1, we might guess that the slope of the tangent line is $m=2$. Using the point-slope formula we can determine the equation of the tangent line

$$ \begin{align*} y - y_1 &= m(x-x_1) \\ y - 1 &= 2(x-1) \\ y &= 2x-1 \end{align*} $$

We should make sure that the line $y=2x-1$ and the curve $y=x^2$ have only the point $(1,1)$ in common.

$$ \begin{align*} 2x-1 &= x^2 \\ x^2 - 2x + 1 &= 0 \\ (x-1)^2 &= 0 \\ x-1 &= 0 \\ x &= 1 \\ y &= 1^2 = 1 = 2(1)-1\ {\color{green}\Large{\checkmark}} \end{align*} $$

There were several steps to this solution. It also involved performing a lot of computations on a piece of paper that I didn't show you. How can we streamline this process? Is there a process that will reliably give us the slope of the tangent line even for every complicated function?

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