The document provides an outline and examples for graphing functions. It includes a checklist for graphing a function, which involves finding where the function is positive, negative, zero or undefined. It then discusses finding the first and second derivatives to determine monotonicity and concavity. Examples are provided to demonstrate this process, including graphing the function f(x) = x + √|x| and f(x) = xe-x^2. Key aspects like asymptotes, points of non-differentiability, and putting the analysis together into a graph are also covered.
1. Section 4.4
Curve Sketching II
V63.0121, Calculus I
April 1, 2009
Announcements
. . . . . .
2. Graphing Checklist
To graph a function f, follow this plan:
0. Find when f is positive, negative, zero, not defined.
1. Find f′ and form its sign chart. Conclude information about
increasing/decreasing and local max/min.
2. Find f′′ and form its sign chart. Conclude concave
up/concave down and inflection.
3. Put together a big chart to assemble monotonicity and
concavity data
4. Graph!
. . . . . .
3. Outline
More Examples
Points of nondifferentiability
Horizontal asymptotes
Vertical asymptotes
Trigonometric and polynomial together
. . . . . .
5. Example
√
|x|
Graph f(x) = x +
This function looks strange because of the absolute value.
But whenever we become nervous, we can just take cases.
First, look at f by itself. We can tell that f(0) = 0 and that
f(x) > 0 if x is positive. Are there negative numbers which
are zeroes for f? Yes, if x = −1 then
√ √
x + |x| = −1 + 1 = 0. No other zeros exist.
. . . . . .
6. Asymptotic behavior
Asymptotically, it’s clear that lim f(x) = ∞, because both
x→∞
terms tend to ∞.
What about x → −∞? This is now indeterminate of the form
−∞ + ∞. To resolve it, first let y = −x to make it look more
familiar:
( √) √ √
lim x + −x = lim (−y + y) = lim ( y − y)
y→∞ y→∞
x→−∞
Now multiply by the conjugate:
√
y − y2
y+y
√
lim ( y − y) · √ = −∞
= lim √
y + y y→∞ y + y
y→∞
. . . . . .
7. The derivative
First, assume x > 0, so
d( √) 1
f′ (x) = x+ x =1+ √
dx 2x
This is always positive. Also, we see that lim f′ (x) = ∞ and
x→0+
′
lim f (x) = 1. If x is negative, we have
x→∞
d( √) 1
f′ (x) = x + −x = 1 − √
dx 2 −x
√
Again, this looks weird because −x appears to be a negative
number. But since x < 0, −x > 0.
. . . . . .
8. Monotonicity
We see that f′ (x) = 0 when
√
1 1 1 1
1= √ =⇒ −x = =⇒ −x = =⇒ x = −
2 4 4
2 −x
Notice also that lim f′ (x) = −∞ and lim f′ (x) = 1. We can’t
x→−∞
x→0−
make a multi-factor sign chart because of the absolute value, but
the conclusion is this:
.′ (x)
f
0 −∓ .
.. . . ∞
. .
+ +
↗ −4 ↘ 0 ↗
. .1 . .
. . f
.(x)
. max cusp
. . . . . .
9. Concavity
If x > 0, then
( )
d 1 1
′′
1 + x−1/2 = − x−3/2
f (x) =
dx 2 4
This is negative whenever x < 0. If x < 0, then
( )
d 1 1
′′ −1/2
= − (−x)−3/2
1 − (−x)
f (x) =
dx 2 4
which is also always negative for negative x. Another way to
1
write this is f′′ (x) = − |x|−3/2 . Here is the sign chart:
4
f′′
. . (x)
−
.− −.
.∞ −
.−
.
. .
⌢ ⌢
0
. f
.(x)
. . . . . .
10. Synthesis
Now we can put these things together.
.′ (x)
f
0 −∓ .
.. . . ∞
. .
+ +
↗ −4 ↘ 0 ↗
. 1. .
. . m
.′′ onotonicity
f
. (x)
−
.− −. .
.−∞
− −
.−
. .. .
⌢ ⌢0 ⌢ c
. oncavity
.1 f
.(x)
0
.. 0
..
4.
. 1. . .
− 0
.1 s
. hape
−
. .4
. .
zero max cusp
. . . . . .
11. Graph
f
.(x)
.−1, 1)
( 44
. −1, 0) .
(
. . x
.
. 0, 0)
(
.1 f
.(x)
0
.. 0
..
4.
. 1. . .
− 0
.1 s
. hape
−
. .4
. .
zero max cusp
. . . . . .
19. Step 0
Find when f is positive, negative, zero, not defined.
. . . . . .
20. Step 0
Find when f is positive, negative, zero, not defined. We need to
factor f:
x+1
1 1
f(x) = + 2 = .
x2
x x
This means f is 0 at −1 and has trouble at 0. In fact,
x+1
= ∞,
lim
x→0 x2
so x = 0 is a vertical asymptote of the graph.
. . . . . .
21. Step 0
Find when f is positive, negative, zero, not defined. We need to
factor f:
x+1
1 1
f(x) = + 2 = .
x2
x x
This means f is 0 at −1 and has trouble at 0. In fact,
x+1
= ∞,
lim
x→0 x2
so x = 0 is a vertical asymptote of the graph. We can make a sign
chart as follows:
−
. .
0
.. +
. x
. +1
−
.1
. .
0
..
+ +
.2
x
0
.
−
. .. . .
∞
0+ .. +
f
.(x)
− 0
.
.1
. . . . . .
24. Monotonicity
We have
x+2
1 2
f′ (x) = − − 3 =− 3 .
2
x x x
The critical points are x = −2 and x = 0. We have the following
sign chart:
−
. .
0
..
+ . −
. (x + 2)
−
.2
−
. .
0
.. +
.3
x
0
.
. . . . . .
25. Monotonicity
We have
x+2
1 2
f′ (x) = − − 3 =− 3 .
2
x x x
The critical points are x = −2 and x = 0. We have the following
sign chart:
−
. .
0
..
+ . −
. (x + 2)
−
.2
−
. .
0
.. +
.3
x
0
.
.′ (x)
f
− −
. . .
∞
0
.. ..
+
−
.2 0
. f
.(x)
. . . . . .
26. Monotonicity
We have
x+2
1 2
f′ (x) = − − 3 =− 3 .
2
x x x
The critical points are x = −2 and x = 0. We have the following
sign chart:
−
. .
0
..
+ . −
. (x + 2)
−
.2
−
. .
0
.. +
.3
x
0
.
.′ (x)
f
− .. −
. . .
∞
0 ..
+
−
↘ .2 0
.
. f
.(x)
. . . . . .
27. Monotonicity
We have
x+2
1 2
f′ (x) = − − 3 =− 3 .
2
x x x
The critical points are x = −2 and x = 0. We have the following
sign chart:
−
. .
0
..
+ . −
. (x + 2)
−
.2
−
. .
0
.. +
.3
x
0
.
.′ (x)
f
− .. −
. . .
∞
0 ..
+
−
↘ .2 ↗ 0
.
. . f
.(x)
. . . . . .
28. Monotonicity
We have
x+2
1 2
f′ (x) = − − 3 =− 3 .
2
x x x
The critical points are x = −2 and x = 0. We have the following
sign chart:
−
. .
0
..
+ . −
. (x + 2)
−
.2
−
. .. .
0+
.3
x
0
.
.′ (x)
f
− .. ∞−
. . .. .
0 +
−
↘ .2 ↗ 0↘
..
. . f
.(x)
. . . . . .
29. Monotonicity
We have
x+2
1 2
f′ (x) = − − 3 =− 3 .
2
x x x
The critical points are x = −2 and x = 0. We have the following
sign chart:
−
. .
0
..
+ . −
. (x + 2)
−
.2
−
. .. .
0+
.3
x
0
.
.′ (x)
f
− .. ∞−
. . .. .
0 +
−
↘ .2 ↗ 0↘
..
. . f
.(x)
m
. in
. . . . . .
30. Monotonicity
We have
x+2
1 2
f′ (x) = − − 3 =− 3 .
2
x x x
The critical points are x = −2 and x = 0. We have the following
sign chart:
−
. .
0
..
+ . −
. (x + 2)
−
.2
−
. .. .
0+
.3
x
0
.
.′ (x)
f
− .. ∞−
. . .. .
0 +
−
↘ .2 ↗ 0↘
..
. . f
.(x)
m
. in V
.A
. . . . . .