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
