## Monday, April 30, 2007

### Precalculus 12.3 The Tangent Line Problem

12.3 The Tangent Line Problem

The Slope of the graph of a function can be used to analyze rates of change at particular points on the graph.

The Slope of a line indicates the rate at which a line rises or falls.
- For a straight line, this rate (or slope) is the same at every point on the line.
- For other graphs than lines, the rate at which the graph rises or falls changes from point to point.

Recall: slope formula = (the change in y's)/(the change in x's) = Δy/Δx

Review: With circles, you were taught about tangent lines. A tangent line is a line that intersects a circle in exactly one point, called the point of tangency. Using this process, to determine the rate at which a graph rises or falls at a single point, you can find the slope of the tangent line at that point.

The tangent line to a graph of a function "f" at a point P(x1, y1) is the line that best approximates the slope of the graph at that point.

Example1: f(x) = x2 - 2x + 1

find the tangent line at the point (1, f(x))

f(1) = 12 -2(1) + 1 = 0
so the tangent line at (1,0) is a straight line and the equation of the tangent line at (1, f(x)) is y=0.

A more precise method then "eyeballing" the tangent line
is making use of the secant line through the point of tangency
and a second point on the line where (x, f(x)) is the first point
and (x + h, f(x + h)) is a second point on the graph of "f", the
slope of the secant line through these two points is:

Msecant = (f(x + h) - f(x))/h

Using the limit process, you can find the
exact slope of the tangent line at (x, f(x)).

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:

M =

lim msec =
h®0

lim (f(x+h) - f(x))/h
h®0

provided the limit exists!

Example 2: given f(x) = 2x + 5, find the slope of the tangent line at (-1, -3)

Msec = (f(x+h) - f(x))/h

= (f(-1 + h) - f(-1))/h
= (2(-1 + h) + 5 - (2(-1) + 5))/h
= (-2 + 2h + 5 +2 - 5)/h
= (2h)/h
= 2
The graph has a slope of 2 at the point (-1, -3)
To find the equation of the tangent line:
y = mx + b
(-3) = 2(-1) + b
-1 = b

so the equation of the tangent line at point (-1, -3) is
y = 2x -1

Example 3: given f(x) = 10x - 2x2 at point (3, 12)

Msec = [10(x+h) - 2(x + h)2 - (10x - 2x2)]/h

Msec = [10x + 10h -2(x2 +2xh + h2) -10x + 2x2]/h

Msec = [10h - 2x2 - 4xh - 2h2 + 2x2]/h

Msec = [10h - 4xh - 2h2]/h

Msec = 10 - 4x - 2h

Slope of the tangent line = M =
lim Msec =
h®0

lim 10 - 4x - 2h = 10 - 4x so at point (3, 12), M = 10 - 4(3) = -2
h®0

Therefore the slope of the tangent line at the point (3, 12) is -2.

To find the equation of the tangent line, use the point and the slope:

y = mx + b
12 = (-2)(3) + b
12 = -6 + b
18 = b

y = 2x + 18 is the equation of the tangent line at the point (3, 12)

Finding a Formula for the slope of a Graph:
Example4: g(x) = x3 at points (1, 1) and (-2, -8)

Msec = [(x + h)3 - (x3)]/h

Msec = [x3 + 3x2h + 3xh2 + h3 - x3]/h

Msec = [3x2h + 3xh2 + h3]/h

Msec = 3x2 + 3xh + h2 , where h¹0

Next take the limit of Msec as h approaches 0.

The slope of the tangent line M =

lim 3x2 + 3xh + h2 = 3x2
h®0

Now use this M = 3x2 for the slope at (1,1)

M = 3(1)2 = 3

now y = mx + b
1 = 3(1) + b
b = -2

so the equation of the tangent line at point (1,1) is y = 3x - 2

Again now at point (-2, -8)

Now use this M = 3x2 for the slope at (-2,-8)

M = 3(-2)2 = 3(4) = 12

now y = mx + b
-8 = 12(-2) + b
-8 = -24 + b
b = 16

so the equation of the tangent line at point (-2,-8) is y = 12x +16

The formula that you derived from the function f(x) = x3 and used the limit process to get M = 3x2 represents the slope of the graph of "f" at the point (x, f(x)).

The Derived function is called the derivative of f at x. It is donoted by f ' (x), which is reas as "f prime of x".

Definition of the Derivative:

The Derivative of "f" at "x" is

f ' (x) = limh®0 (f(x+h) - f(x))/h

provided the limit exists!

Example 5: f(x) = x2 - 3x + 4

f ' (x) = limh®0 [(x + h)2 - 3(x + h) + 4 - (x2 - 3x + 4)]/h

f ' (x) = limh®0 [x2 + 2xh + h2 - 3x -3h + 4 - x2 + 3x -4]/h

f ' (x) = limh®0 [2xh + h2 - 3h]/h

f ' (x) = limh®0 2x + h - 3 = 2x - 3

f '(x) = 2x - 3

So therefore the derivative of f(x) = x2 - 3x + 4 is f ' (x) = 2x - 3

NOTE that in addition to f ' (x) , other notations can be used to denoted the derivatives of y = f(x). The most common are:

(dy)/(dx)

y'

(d/(dx))[f(x)]

and Dx[y]

Using the derivative of f(x) = x3 - x, find the slope of the tangent line at point (2,6).

f'(x) = limh®0 [(x + h)3 - (x + h ) - (x3 - x)]'h

f'(x) = limh®0 [x3 + 3x2h + 3xh2 + h3 - x - h - x3 - x]/h

f'(x) = limh®0 [3x2h + 3xh2 + h3 - h]/h

f'(x) = limh®0 (3x2 + 3xh + h2 - 1)

f ' (x) = 3x2 - 1

So at the point (2, 6), the slope is
f ' (2) = 3(2)2 - 1 = 3(4) - 1 = 12 - 1 = 11

So the equation of the tangent line at point (2,6) is
6 = 11(2) + b
6 = 22 + b
b = -16

so the equation of the tangent line at point (2,6) is y = 11x - 16

Example 6: Find the derivative of f. Use the derivative to determine any points on the graph of f where the tangent line is horizontal.

f(x) = x3 - 3x

f ' (x) = limh®0 [(x + h)3 - 3(x + h) - (x3 - 3x)]/h

f ' (x) = limh®0 [x3 + 3x2h + 3xh2 + h3 - 3x -3h - x3 + 3x]/h

f ' (x) = limh®0 [3x2h+ 3xh2 + h3 -3h]/h

f ' (x) = limh®0 [3x2 + 3xh+ h2 -3]

f ' (x) = 3x2 - 3

Recall a horizontal line has a slope of zero so

f ' (x) = 0 so
0 = 3x2 - 3
3 = 3x2
1 = x2
x = ±1

putting these values in for x:

So "f" has horizontal tangents at (1, 0) adn (-1, 0)

If you need extra help, watch the Chapter 2 lesson 1 tutorial at
http://www.calculus-help.com/funstuff/phobe.html