Chapter 1.1b Functions + 1.2a Graphs of Functions
I. Application: A right triangle is formed in the first quadrant by the x-axis and y-axis and a line through the point (2, 1). Write the area of the triangle as a function of “x”, and determine the domain of the function.
1. Graph the triangle:
- one side of the triangle is the x-axis and another side is the y-axis. Since we do not know exactly where the points of the third line cross the 2 axes, we use the points (0,y) and (x,0).
A = ½ (base)(height) = ½ xy
Since (0, y) , (2, 1) and (x, 0) all lie on the same line, the slopes between any pairs of points are equal.
M = (1 – y)/(2 – 0) = (1 – 0)/(2 – x)
1 – y = 2/(2 – x)
y = -2/ ( 2 – x) + 1
y = (-2 + (2 – x))/ (2 – x)
y = -x / (2 – x)
y = x / (x – 2)
Now using substitution:
A = ½ x (x / (x – 2)) = x2/(2x – 4)
The domain is (2, ∞ ) since the area has to be greater than zero.
II. Evaluating a Difference Quotient:
If f (x) = 2x, then using the difference quotient below
Using f(x) = 2x, plug into the equation above:
2h/h = 2
Example 2: Given 5x – x2,
I. Application: A right triangle is formed in the first quadrant by the x-axis and y-axis and a line through the point (2, 1). Write the area of the triangle as a function of “x”, and determine the domain of the function.
1. Graph the triangle:
- one side of the triangle is the x-axis and another side is the y-axis. Since we do not know exactly where the points of the third line cross the 2 axes, we use the points (0,y) and (x,0).
A = ½ (base)(height) = ½ xy
Since (0, y) , (2, 1) and (x, 0) all lie on the same line, the slopes between any pairs of points are equal.
M = (1 – y)/(2 – 0) = (1 – 0)/(2 – x)
1 – y = 2/(2 – x)
y = -2/ ( 2 – x) + 1
y = (-2 + (2 – x))/ (2 – x)
y = -x / (2 – x)
y = x / (x – 2)
Now using substitution:
A = ½ x (x / (x – 2)) = x2/(2x – 4)
The domain is (2, ∞ ) since the area has to be greater than zero.
II. Evaluating a Difference Quotient:
If f (x) = 2x, then using the difference quotient below
Using f(x) = 2x, plug into the equation above:
2h/h = 2
Example 2: Given 5x – x2,
-5 – h , where h cannot equal zero
Example 3:
Increasing, Decreasing, and Constant Functions
A function “f” is increasing on an interval if, for any x1 and x2 in the interval, when x1 is less than x2 implies that f (x1 ) is less than f(x2 )
A function “f” is decreasing on an interval if, for any x1 and x2 in the interval, when x1 is less than x2 implies that f (x1 ) is greater than f(x2 )
A function “f” is constant on an interval if, for any x1 and x2 in the interval, f (x1 ) = f(x2 )
Check out this website:
http://www.mathematicshelpcentral.com/lecture_notes/precalculus_algebra_folder/increasing_and_decreasing_functions.htm
Example 3:
Increasing, Decreasing, and Constant Functions
A function “f” is increasing on an interval if, for any x1 and x2 in the interval, when x1 is less than x2 implies that f (x1 ) is less than f(x2 )
A function “f” is decreasing on an interval if, for any x1 and x2 in the interval, when x1 is less than x2 implies that f (x1 ) is greater than f(x2 )
A function “f” is constant on an interval if, for any x1 and x2 in the interval, f (x1 ) = f(x2 )
Check out this website:
http://www.mathematicshelpcentral.com/lecture_notes/precalculus_algebra_folder/increasing_and_decreasing_functions.htm
Relative Minimum and Relative Maximum:
1. A function value f (a) is called a relative minimum of “f” if there exists an interval that contains a such that:
1. A function value f (a) is called a relative minimum of “f” if there exists an interval that contains a such that:
So you can see that the function is increasing from (- ∞ , -3) and then again from (2, ∞). The function is decreasing from (-3, 2)
Now lets see an example of relative minimum and relative maximum points:
So you can see that the function's relative minimum point from ( - ∞, 0 ) is
"a" while its relative minimum point from ( 0, ∞) is "c" . The relative maximum point is "b".
