Transcript Graphingx

GRAPHING
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To graph a function it will be important to find the high points and low points in
the graph as well as where the function is increasing and decreasing. The high
points and low points are called the RELATIVE EXTREMA (relative maximums
and relative minimums). The function will change from increasing to decreasing
and vice versa at those extrema points. Therefore, the first goal is to find where
the relative maximums and minimums are possible. The graph below shows a
function with a relative maximum at the point (-3,25) and a relative minimum at
(1.5,-25). Notice that the function (from left to right) increases to the maximum
point, decreases between the maximum point and the minimum point, then
increases past the minimum point. The values of x where the relative extrema
occur are 1.5 and – 3. These values of x are called CRITICAL NUMBERS.
Notice that at the maximum and minimum points that the tangent line would be
horizontal there. The slopes of those tangent lines would be zero!!!!
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Definition: 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:
f ( x)  3 x 3  9 x 2  4
has a domain of all real numbers and its 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?
FACTS:
 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,
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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.
And remember: A function will change from
increasing to decreasing and vice versa at critical
numbers and vertical asymptotes.
TO
FIND WHERE THE FUNCTION IS INCREASING AND
DECREASING:
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Place the critical numbers on a number line.
Pick a number on each side of each critical number and
plug into the first derivative.
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:
Example:
f ( x)  3 x 3  9 x 2  4
has
f ' ( x)  9 x 2  18x  9 x( x  2)
And critical numbers at x = 0 and x = 2.
f’(-1) = -9(-3) = +27
f’(-1) = 9(-1) = -18
f’(3) = 27(1) = +27
increasing
decreasing
increasing
up to x = 0
between x = 0 and 2
from x = 2 on
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0
2
EX:
f ( x)  3 x 3  9 x 2  4
WITH
f ' ( x)  9 x 2  18x  9 x( x  2)
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So the intervals where this function increases are (,0)  (2, ) .
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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.
EX:
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f ( x)  3 x 3  9 x 2  4
WITH
f ' ( x)  9 x 2  18x  9 x( x  2)
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.
EX:
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f ( x)  3 x 3  9 x 2  4
WITH
f ' ( x)  9 x 2  18x  9 x( x  2)
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.
f ( x)  3 x  9 x  4
3
2
CONCAVITY
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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.
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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.
DEFINITION: PIN numbers are values of x in the
domain of the function that make the SECOND
derivative zero or undefined.
(These are like critical numbers for the second derivative).
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Back to our example:
f ( x)  3 x 3  9 x 2  4
f ' ( x)  9 x 2  18x  9 x( x  2)
f ( x)  18x  18  18( x  1)
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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”
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1
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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.
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Here is another example of how to find critical numbers and where a
function is increasing and decreasing, concave up and concave down:
Example:
f ( x)  x 3  9 x 2
has a first derivative of
f ' ( x)  3x 2  18x  3x( x  6)
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So the critical numbers are x = 0 and x = 6.
When I check to see where the first derivative is positive and negative, (where the
function is increasing and decreasing respectively),
f’(-1)= +18
f’(3)= - 27
f’(7)= +21
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0
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6
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).
Remember the y – values of 0 and 108 come from plugging the corresponding x –
value into the original function
f ( x)  x 3  9 x 2
f ( x)  x 3  9 x 2
Remember from the previous slide:
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I now look at the second derivative which is:
f ' ( x)  3x 2  18x  3x( x  6)
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.
f’’(0)= - 18
f ‘’(5)= +12
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3
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Therefore, the function is concave down (frowning at me) up to 3 and concave
up (smiling at me) on the right side of 3.
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Since the concavity changed at 3, there is a point of inflection at (3,-64). The
– 64 is found by plugging x = 3 into
f ( x)  x 3  9 x 2
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Therefore, there is a maximum point at (0,0) and a minimum point at (6,-108).
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.
Since the concavity changed at 3, there is a point of inflection at (3,-64).
f ( x)  x 3  9 x 2
Max at (0,0)
Inflection point at (3, -63)
Min at (6, -108)