Hybridoma Technology ( Production , Purification , and Application )
Lesson 21: Curve Sketching II (Section 4 version)
1. Section 4.4
Curve Sketching II
V63.0121, Calculus I
April 2, 2009
.
.
Image credit: svenwerk
. . . . . .
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
Logarithmic
. . . . . .
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
. . . . . .