Download Business Calculus I

Business Calculus I
Graphing Explanation Worksheet
Critical numbers are values of x in the domain of a function that make the first derivative zero or
undefined. So, critical numbers are x values where the tangent line is horizontal or vertical. Critical
numbers are also the values of x where the function’s graph could reach a maximum (high point) or a
minimum (low point).
Example:
(When you see “  ” refer back this function) f ( x)  3x 3  9 x 2  4 has a domain of all real
numbers and his first derivative is: f ' ( x)  9 x 2  18x  9 x( x  2) . The critical numbers are x = 0 and
x = 2. So the function may have a low point or high point at x = 0 or x = 2. How do you find out?
When the first derivative is positive, the slopes of the tangent lines are positive. If the slope of a line is
positive it slopes up when you graph it. If the tangent line is sloping up, then the function is increasing.
Likewise, when the first derivative is negative, the slopes of the tangent lines are negative. If the slope of
a line is negative the line slopes down when you graph it. If the tangent line is sloping down, then the
function is decreasing. A function will change from increasing to decreasing and vice versa at critical
numbers and vertical asymptotes.
1. Place the critical numbers on a number line.
2. Pick a number on each side of each critical number and plug into the first derivative.
3. If the first derivative is a positive number, the function is increasing on that interval or section of
the number line. If the first derivative is a negative number, the function is decreasing on that
particular interval. See below:
f ' ( 1) = -9(-3) = +27
f ' (3) = 27(1) = +27
f ' (1) = 9(-1) = -18
increasing
decreasing
increasing
up to x = 0
between x = 0 and 2
from x = 2 on
______________________________________________
\
\
0
2
So the intervals where this function increases are (,0)  (2, ) . The interval where this function
decreases is (0,2). In other words (from left to right), the graph of this function does the following:
GOES UP TO SOME POINT, GOES DOWN TO SOME POINT, THEN GOES UP!!
That should help you to start a mental image of this function. The points where the changes occur are the
maximum and minimum points. Let’s find them.
The function first increases up to x = 0 then decreases. Picture that in your mind. Isn’t that a high point?
A maximum? YES. So this function has a maximum point at x = 0. The point itself would be (0,4). I
found the four from plugging x = 0 into the original function  . The original function always will give
me my y values.
The function then continues to decrease until x = 2 then increases. Picture that in your mind. Isn’t that a
low point? A minimum? YES. So this function has a minimum point at x = 2. The entire point would
be (2,-8). Again I found the eight by plugging x = 2 into the original function  . Now sketch the picture.
We can make sure our picture is even more accurate by using the second derivative of the function to look
at the concavity of the function. Concavity can be either up or down. In the sketch above, the function is
concave down (“frowning”) from the left hand side to somewhere close to x = 1. Then the function
becomes concave up (“smiling” ). The second derivative will tell you where the function is concave up
and down and where the concavity changes very much like the first derivative tells you where the
function is increasing and decreasing.
When the second derivative is positive, the function is “smiling” (concave up).
When the second derivative is negative, the function is “frowning” (concave down).
The function will change from concave up to down and vice versa at certain values of x called PIN
numbers or possibly at vertical asymptotes. PIN numbers for the second derivative are the same as
critical numbers for the first derivative.
Defn: PIN numbers are values of x in the domain of the function that make the SECOND derivative zero
or undefined.
Watch this: f ( x)  18x  18  18( x  1) . So the PIN number is x = 1. Place the PIN number on the
number line and test what happens to the second derivative on either side of x = 1.
f (0)  18(1)  18
f (2)  18(1)  18
Negative means
Positive means
“frowning”
“smiling”
____________________________________
/
x=1
That matches what our sketch shows. Since the concavity changed at x = 1, the function has what is
called a point of inflection. To find that point of inflection plug x = 1 back into the original function  .
This will give you a y value of y = -2. So the POINT of inflection is (1,-2). You could then label that
point in the sketch.
Here is another example of how to find critical numbers and where a function is increasing and
decreasing, concave up and concave down:
f ( x)  x 3  9 x 2 has a first derivative of f ' ( x)  3x 2  18x  3x( x  6) . So the critical numbers are x = 0
and x = 6. When I check to see where the first derivative is positive and negative, ie, where the function
is increasing and decreasing respectively, I find the function is increasing from negative infinity to 0 and
from 6 to positive infinity. The function is decreasing between 0 and 6. Therefore, there is a maximum
point at (0,0) and a minimum point at (6,-108).
I now look at the second derivative which is f ' ' ( x)  6 x  18  6( x  3) . The PIN number for the second
derivative is x = 3. When I test to see what happens with the second derivative on either side of 3, this is
what I find. The second derivative is negative on the left side of 3---from negative infinity up to 3, and
positive on the right side of 3---from 3 to positive infinity. Therefore, the function is concave down
(frowning at me) up to 3 and concave up (smiling at me) on the right side of 3. Since the concavity
changed at 3, there is a point of inflection at (3,-64).
Notice: The way I found the y-value for all of the points (max, min, and inflection) was to plug the xvalue into the original function f ( x)  x 3  9 x 2 .