Sophisticated Graphing

Week 8.
Sophisticated Graphing
We will combine pre-calculus and calculus methods to sketch much more accurate
graphs than in your previous classes. First we need to learn some applications of the
derivative.
8.1. The First Derivative and the Shape of a Graph
Definition 8.1.
• A function f
of f .
• A function f
of f .
• A function f
• A function f
has an absolute maximum at c if f (c) ≥ f (x) for all x in the domain
has an absolute minimum at c if f (c) ≤ f (x) for all x in the domain
has a local maximum at c if f (c) ≥ f (x) when x is near c.
has a local minimum at c if f (c) ≤ f (x) when x is near c.
The absolute maximum and minimum are sometimes called global extreme values of
f . The local max and min are sometimes called relative extreme values. Notice that
4
2
some functions do not
√ have all of these values. For instance f (x) = x − 4x has global
minimums (at x = ± 2), but no global maximum.
Functions like f (x) = x or f (x) = x3 have neither global nor local extreme values at all.
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Calculus 150AL
Theorem 8.1. (The Extreme Value Theorem) If f is continuous on a closed interval
[a, b], then f attains an absolute maximum and an absolute minimum somewhere on that
interval.
Example 8.1. Identify the absolute extrema and relative extrema for the function
f (x) = x2 on the interval [−1, 2].
Solution. First of all, from the extreme value theorem, we know that there are is an
absolute maximum and an absolute minimum. Extreme values of a function must have
derivative equal to zero. Here are the steps to follow:
(1) Differentiate f (x): f (x) = 2x
(2) Now set f (x) = 0 and solve for x. So the only time when f (x) = 0 is when x = 0.
This is called a critical point of the function.
(3) Because we have a closed interval, we also need to check the end points, f (−1)
and f (2).
(4) Check the values at the end points and critical points. An endpoint can only be
an absolute extreme value, not a relative one. A critical point could be either (or
both).
f (0) = 0
f (−1) = 1
f (2) = 4
So we compare the values and see that there is an absolute max at x = 2 and an
absolute (and local) min at x = 0. There are no local maxima.
Your Turn. † Identify the absolute extrema and relative extrema for the function f (x) =
2t3 + 3t2 − 12t + 4 on [−4, 2].
Theorem 8.2. (Increasing/Decreasing Test)
(1) If f (x) > 0 on an interval, then f is increasing on that interval.
(2) If f (x) < 0 on an interval, then f is decreasing on that interval.
Week 8. Sophisticated Graphing
Theorem 8.3. (First Derivative Test) Suppose c is a
function f .
(1) If f changes from positive to negative at c, then f
(2) If f changes from negative to positive at c, then f
(3) If f does not change sign at c, then f has no local
45
critical value of a continuous
has a local maximum at c.
has a local minimum at c.
minimum or minimum at c.
Example 8.2. Find the intervals where the following function is increasing and decreasing. Also, find the local maxima and minima if they exist.
5
40
f (x) = −x5 + x4 + x3 + 5
2
3
Solution.
• Differentiate the function to learn about its increasing and decreasing
intervals.
f (x) = −5x4 + 10x3 + 40x2
• Factor and solve for f (x) = 0 to obtain the critical points:
0 = −5x2 (x − 4)(x + 2) therefore x = 0, −2, 4
These split up the domain of f (x) (which is all real numbers) into the intervals
(−∞, −2], [−2, 0], [0, 4], [4, ∞).
• point in each interval, so that we can apply the Increasing/Decreasing test above.
Remember that the most important thing here is the sign of the answer, not the
value.
f (−3) ≤ 0 =⇒ f is decreasing on (−∞, −2]
f (−1) ≥ 0 =⇒ f is increasing on [−2, 0]
f (1) ≥ 0 =⇒ f is increasing on [0, 4]
f (5) ≤ 0 =⇒ f is decreasing on [4, ∞)
• Using this information, we can apply the First Derivative Test to say that f (−2)
is a local minimum since the interval below it is decreasing and the interval above
it is increasing. Following the theorem, we see that f (0) is neither, and f (4) is a
local maximum.
• Finally, use all the information from the previous parts to draw a detailed graph.
Your Turn. † Find the intervals where the following function is increasing and decreasing.
Also, find the local maxima and minima if they exist.
f (x) = 4x3 + 3x2 − 6x + 1
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Calculus 150AL
8.2. The Second Derivative and the Shape of a Graph
Definition 8.2.
• If the graph of f lies above all its tangents on an interval I, then f is called concave
upward on I. It looks like a cup.
• If the graph of f lies below all its tangents on an interval I, then f is called concave
downward on I. It looks like an upside down cup.
Every continuous part of a graph is either concave up, concave down, or constant.
Definition 8.3. A point P on a curve y = f (x) is called an inflection point if f is
continuous there and the curve changes concavity at P . Any point where the second
derivative changes sign is an inflection point.
Theorem 8.4. (Concavity Test)
(1) If f (x) > 0 on an interval, then f is concave upward on that interval.
(2) If f (x) < 0 on an interval, then f is concave downward on that interval.
Theorem 8.5. (Second Derivative Test) Suppose f is continuous near c.
(1) If f (c) = 0 and f (c) > 0, then f has a local maximum at c.
(2) If f (c) = 0 and f (c) < 0, then f has a local minimum at c.
(3) If f (c) = 0 and f (c) = 0, then this test is inconclusive.
Example 8.3. Discuss the concavity and points of inflection of the graph of the
following function. Also apply the Second Derivative Test to confirm our findings about
the local maxima and minima.
y = x4 − 4x3
Solution.
• Differentiate the function twice to learn about its concavity.
f (x) = 4x3 − 12x2
f (x) = 12x2 − 24x
• Factor and solve for f (x) = 0 to obtain the inflection points:
0 = 12x(x − 2) therefore x = 0, 2
Week 8. Sophisticated Graphing
47
These split up the domain of f (x) (which is all real numbers) into the intervals
(−∞, 0], [0, 2], [2, ∞).
• Check a point in each interval, so that we can apply the concavity test above.
Remember that the most important thing here is the sign of the answer, not the
value.
f (−1) ≥ 0 =⇒ f is concave up on (−∞, 0]
f (1) ≤ 0 =⇒ f is concave down on [0, 2]
f (3) ≥ 0 =⇒ f is concave up on [2, ∞)
• We can apply the Second Derivative Test to find local maxs and mins. We want to
examine points where the first derivative of f is equal to 0. So we need to factor
and solve f (x) = 0
0 = 4x2 (x − 3) therefore x = 0, 0, 0, 3
We check: f (0) = 0, so it is inconclusive at this point. But we do get that
f (3) > 0 implies x = 2 is a minimum of the graph. At this point we could use
the first derivative test to figure out if x = 0 is (or isn’t) a max or min of f .
Your Turn.
∗∗
f (x) = x5 − 2x3 + x
• Find increasing and decreasing intevals.
• Discuss the concavity and points of inflection of the graph of the following function.
• Also apply the Second Derivative Test to confirm our findings about the local
maxima and minima. If the test is inconclusive, apply the First Derivative Test.
8.3. Infinite Limits: Vertical Asymptotes
Definition 8.4. If f is defined on both sides of a (but possibly not at x = a), and the
values of f (x) can be made arbitrarily large by taking x sufficiently close to a, but not
equal to a, then we write
lim f (x) = ∞
x→a
There are also corresponding definitions for the limit equaling negative infinity, and for
left- and right-hand limits which are infinite. You can write these out or look them up as
an exercise.
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Calculus 150AL
Definition 8.5. If f (x) has an infinite limit (from left or right or both) as x approaches
a, then the line x = a is a vertical asymptote of f (x).
Example 8.4.∗∗ Does f (x) =
1
have any vertical asymptotes?
x2
Solution. We should check the left and right-handed limits separately. Unfortunately,
we don’t have an exact way (yet) to calculate or show that certain limits are infinte. We
need to think about this intuitively.
The numerator of this fraction is constant. The denominator is x2 . As we approach
zero from either side, we get smaller and smaller positive numbers in the denominator.
Dividing by smaller and smaller numbers gives us bigger and bigger numbers. Since we
can get as close to zero as we want without actually reaching it (.0001, .00001, etc), then
1
1
when we divide, we get arbitrarily large numbers. Therefore lim 2 = lim 2 = ∞. By
x→0− x
x→0+ x
the previous theorem, since the left and right limits are the same, the limit exists and is
also equal to ∞.
1
=∞
x→0 x2
lim
Example 8.5. Convince yourself that the following are true:
1
= −∞
x→0− x
1
=∞
(2) lim
x→0+ x
1
(3) lim = undefined
x→0 x
1
= undefined
(4) lim
x→1 x − 1
(1) lim
2x
exist? If so what is its value?
x→3 x − 3
Your Turn. Does lim
Week 8. Sophisticated Graphing
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8.4. Limits at Infinity: Horizontal Asymptotes
Now we will talk about “end-behavior” of functions, or what happens as x approaches
positive (or negative) infinity. The notation is what you would expect:
lim f (x) = L
x→+∞
Definition 8.6. If lim f (x) = L exists, then the line y = L is a horizontal asymptote
x→+∞
of f (x).
Example 8.6.∗ Find the limit
x2
x→∞ 5 − x3
Solution. It’s especially helpful when you are doing these problems to remember that as
1
x approaches infinity, then n approaches 0, for all n > 0. As the bottom gets bigger,
x
then expression itself gets smaller and smaller, and non-fractional exponents only speed
up the process.
The standard trick is to divide both numerator and denominator by the largest exponent
of x (in this example, x3 )
lim
x2
x2 /x3
1/x
0
=
= 0.
=
lim
= lim
3
3
3
3
3
x→∞ 5 − x
x→∞ 5/x − x /x
x→∞ 5/x − 1
0−1
lim
Your Turn.
∗
Find: lim
x→−∞
t
t−5
8.5. Summary of the method
(1) Precalculus analysis (See lecture 1)
(a) Find the domain and range.
(b) Look for symmetry.
(c) Find X and Y -intercepts.
(2) Calculus analysis:
(a) First derivative: Find increasing and decreasing intervals. Find local/global
min and max if they exist.
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Calculus 150AL
(b) Second derivative: Check intervals for concavity and inflection points.
(c) Limits at infinity and infinite limits: Check for asymptotes. (See sections 2.2
and 2.3)
(3) Plot important points. Sketch the graph. Connect the points with a smooth curve
as to maintain all the properties found in steps 1 and 2.
Example 8.7. Sketch the graph of the function
2
f (x) = √ + x
x
Solution. These types of problems can be very long! I’m leaving most of the work in
this example, to try to keep it short.
Precalculus analysis
(1) The domain is x > 0, the range is y > 0
(2) This function is neither even nor odd.
(3) There are no X or Y -intercepts
Calculus analysis
(1) Take the first derivative of f :
1
1
x3/2 − 1
f (x) = − 2x−3/2 + 1 = − 3/2 + 1 =
2
x
x3/2
Set this equal to zero to find the critical points of f :
x3/2 − 1
= 0 =⇒ x3/2 = −1 =⇒ x = 1
x3/2
Now check the sign of each of the intervals by testing points. f ( 12 ) ≈ −1.83 so
f (x) < 0 on 0 < x < 1. If you are on a quiz or test you won’t have a calculator,
but you can still work this out (mentally or as scratch work is fine). I’m using
(+) =“positive” and (−) = “negative,” and the reasoning goes something like this:
f ( 12 ) =
((+), less than 1)3/2 − 1
((+), less than 1) − 1
(−)
=
=
= (−)
3/2
(+)
(+)
(+)
For the interval x > 1, I’ll choose x = 4. The work is similar, but with the opposite
result:
((+), more than 1)3/2 − 1
(−)
((+), more than 1) − 1
=
= (+)
=
f (4) =
3/2
(+)
(+)
(+)
Because f (x) < 0 on 0 < x < 1, then f is decreasing on (0, 1), and because
f (x) > 0 on x > 1, then f is increasing on (1, ∞). The first derivative test tells
us that f has a minimum at x = 1:
⎫
f (x) < 0 for x < 1 ⎪
⎬
⇒ MIN at x = 1
f (c) = 0,
⎪
⎭
f (x) > 0 for x > 1
Week 8. Sophisticated Graphing
51
(2) Now for the second derivative analysis.
3
3
f (x) = x−5/2 = √
2
2 x5
Set this equal to zero to find inflection points.
3
√ =0
2 x5
But there are no solutions to this equation! So the entire domain (0, ∞) is the
only interval we need to check. Use the method in the first derivative analysis to
find the sign of f (x) on the interval (0, ∞). We find that f (x) > 0 for all x > 0.
This means that f (x) is concave up on (0, ∞). In this case we don’t need to do
the second derivative test since we already successfully found the minimum of f
using the first derivative test.
(3) Now use limits to find the vertical and horizontal asymptotes.
Recall from algebra that there is a vertical asymptote at any value of x that
causes division by zero. The only such value here is x = 0. So we know that f (x)
has a vertical asymptote at x = 0. Normally we would check both the limit as x
approaches 0 from the left and separately from the right. However, since the f (x)
is not defined for negative values, you only need to look at the right-hand limit.
1
As x gets very close to 0, then the limit of n for all n > 0 approaches either
x
positive or negative infinity, depending on the value of n. In this case, the square
root of a positive number could be negative or positive, but as a convention, we
use the positive values only. Use the rules: ∞ + ∞ = ∞ and k · ∞ = ∞ for any
constant k.
1
lim+ 1/2 = ∞
x→0 x
2
=⇒ lim+ 1/2 = 2∞ = ∞
x→0 x
2
=⇒ lim+ 1/2 + x = ∞ + ∞ = ∞
x→0 x
You should know how to write this in limit notation:
2
lim+ √ + x = ∞
x→0
x
This says that as x approaches 0 from the right-hand side, then the values of f (x)
approach infinity.
To find horizontal (or oblique) asymptotes, you do need to use limits at infinity.
The equation of the asymptote is given by g(x) = mx + b where,
f (x)
x→∞ x
So in this case,
m = lim
and
b = lim [f (x) − mx]
x→∞
1
m = lim f (x) = lim − 3/2 + 1 = 1
x→∞
x→∞
x
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Calculus 150AL
2
b = lim (f (x) − mx) = lim √ + x − 1 · x = 0
x→∞
x→∞
x
Therefore, the equation of the asymptote is y = 1x + 0 = x
Graphing Sketch in the asymptotes as dotted lines, then plot the minimum at (1, 3).
Now sketch the rest of the graph, maintaining the properties we found. Don’t forget to
think about concavity.
Your Turn. ∗ Make an analysis as suggested in the summary above and sketch the graphs
of
2 2
(t − 1)
(1) h(t) = t√
(2) f (x) = sin x
x2
(3) g(x) =
x−2