Section 12.3

The Tangent Line Problem

Tangent Line to a Graph

In Algebra I you learned that the slope of a line indicated the rate at which it rises or falls.For a line this rate (or slope) is the same at every point.For graphs other than lines, the rate at which the graph rises or falls changes from point to point.

Look at the parabola below

To determine the rate at which a graph rises or falls at a single point, you find the slope of the tangent line at that point.In simple terms, the tangent line to the graph of a function at a point is the line that best approximates the slope of the graph at the point.

In geometry, you learned that a line tangent to a circle intersects the circle in exactly one point.Tangent lines of noncircular graphs can intersect the graph at more than one point.

Slope of a Graph

Because a tangent line approximates the slope of the graph at a point, the problem of finding the slope of a graph at a point is the same as finding the slope of the tangent line at the point.

Example 1Use the figure on the next slide to approximate the slope of the graph of f(x) = x3 at the point (1, 1)

1

3

At (1, 1) this graph appears to have a slope of 3.

Slope and the Limit Process

In example 1 we found the slope of a graph at a particular point by “eyeballing” the tangent line at that point.A more precise method of approximating tangent lines makes use of a secant line through the point of tangency and a second point on the graph.

secm

f x h f xx h x

The slope of the secant line through two points is given by

the right side of this equation is called the difference quotient.

sec

f x h f xchange in ym change in x h

The beauty of this procedure is that you obtain a better approximation of the slope of the tangent line by choosing two points closer and closer to the point of tangency.

Definition of the Slope of a Graph

The slope m of the graph of f at the point (x, f(x)) is equal to the slope of its tangent line at (x, f(x)), and is given by

This definition is a major concept in calculus.

sech 0m lim m

h 0

f x h f xlim h

Example 2Find the slope of the graph of f(x) = x3 at the point (2, 8).

sec

f 2 h f 2m h

3 32 h 2h

2 38 12h 6h h 8h

2h 12 6h hh

212 6h h , h 0

tangent sech 0m lim m

Now take the limit of msec as h approaches 0.

2h 0lim 12 6h h

12

Example 3Find the slope of f(x) = -3x + 5 using the difference quotient.

h 0

f x h f xm lim h

h 0

3 x h 5 3x 5lim h

h 03x 3h 5 3x 5lim h

h 03hlim h

3

h 0lim 3

Let’s look at the difference in Example 2 and 3.In Example 2, you were finding the slope of a graph at a specific point (c, f(c)).

h 0

f c h f cm lim h

slope at a

specific point

In Example 3, you were finding a formula for the slope at any point on the graph. In such cases, you should use x, rather than c, in the difference quotient.

In all nonlinear graphs this will produce a function of x, which can then be evaluated to find the slope at any desired point.

h 0

f x h f xm lim h

formula for

slope

Example 4(a)Find a formula for the slope of

the graph of f(x) = x2 – 2. (b)Then find the slopes at the

points (-3, 7) and (1, -1).

sec

f x h f xm h

2 2x h 2 x 2h

2 2 2x 2xh h 2 x 2h

22xh hh

2x h, h 0

a.

h 0

m lim 2x h

2x

b. Now find the slope at (-3, 7) and (1, -1).The slope at (-3, 7) is -6.The slope at (1, -1) is 2.

End of 1st Day’s Notes

In the last example from yesterday you started with f(x) = x2 – 2 and used the limit process to derive another function m = 2x, that represents the slope of the graph of f at the point (x, f(x)).This derived function is called the derivative of f at x. It is denoted byf’(x).

Definition of the Derivative

The derivative of f at x is given by

provided this limit exists.

h 0

f x h f xf' x lim h

Example 5Find the derivative of

f(x) = 4x2 – 5x

h 0

f x h f xf' x lim h

2 2

h 0

4 x h 5 x h 4x 5xlim h

2 2 2

h 04x 8xh 4h 5x 5h 4x 5xlim h

2

h 08xh 4h 5hlim h

h 0lim 8x 4h 5

8x 5

There are other notations for a derivative that you will see in calculus. They are

dydx y'

d f xdx xD y

Example 6

Find f' x for f x x 1. Then find the slopes of the graph of f at the points 4, 3 and 9, 4 .

h 0

f x h f xf' x lim h

h 0

x h 1 x 1lim h

h 0x h xlim h

We must rationalize the numerator to find the limit.

h 0

x h x x h xf' x lim h x h x

h 0x h xlim

h x h x

h 0hlim

h x h x 12 x

1f' 42 4

The slope at 4, 3 is

The slope at 9, 4 is

14

1f' 92 9

16