Sample questions for test 2 There may be other types of questions, but being able to do these should mean you will do really well on this test.
using Roots, Plots, SymPy
Limits
Graphically identify the following limits or say why they do not exist:
$$~
\lim_{x \rightarrow 0} \frac{\sqrt{x + 25} - 5}{x}
~$$
A: We make the function then plot near 0:
f(x) = (sqrt(x + 25) - 5) / x
plot(f, -1, 1)
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
-3.0
-2.9
-2.8
-2.7
-2.6
-2.5
-2.4
-2.3
-2.2
-2.1
-2.0
-1.9
-1.8
-1.7
-1.6
-1.5
-1.4
-1.3
-1.2
-1.1
-1.0
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
-4
-2
0
2
4
-3.0
-2.8
-2.6
-2.4
-2.2
-2.0
-1.8
-1.6
-1.4
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
y1
0.0960
0.0965
0.0970
0.0975
0.0980
0.0985
0.0990
0.0995
0.1000
0.1005
0.1010
0.1015
0.1020
0.1025
0.1030
0.1035
0.1040
0.1045
0.0965
0.0966
0.0967
0.0968
0.0969
0.0970
0.0971
0.0972
0.0973
0.0974
0.0975
0.0976
0.0977
0.0978
0.0979
0.0980
0.0981
0.0982
0.0983
0.0984
0.0985
0.0986
0.0987
0.0988
0.0989
0.0990
0.0991
0.0992
0.0993
0.0994
0.0995
0.0996
0.0997
0.0998
0.0999
0.1000
0.1001
0.1002
0.1003
0.1004
0.1005
0.1006
0.1007
0.1008
0.1009
0.1010
0.1011
0.1012
0.1013
0.1014
0.1015
0.1016
0.1017
0.1018
0.1019
0.1020
0.1021
0.1022
0.1023
0.1024
0.1025
0.1026
0.1027
0.1028
0.1029
0.1030
0.1031
0.1032
0.1033
0.1034
0.1035
0.1036
0.1037
0.1038
0.1039
0.1040
0.1041
0.096
0.098
0.100
0.102
0.104
0.0962
0.0964
0.0966
0.0968
0.0970
0.0972
0.0974
0.0976
0.0978
0.0980
0.0982
0.0984
0.0986
0.0988
0.0990
0.0992
0.0994
0.0996
0.0998
0.1000
0.1002
0.1004
0.1006
0.1008
0.1010
0.1012
0.1014
0.1016
0.1018
0.1020
0.1022
0.1024
0.1026
0.1028
0.1030
0.1032
0.1034
0.1036
0.1038
0.1040
Near 0, the y value is going towards 0.1.
$$~
\lim_{x \rightarrow 0} \frac{|x|}{x}
~$$
We plot and look:
f(x) = abs(x) / x
plot(f, -1, 1)
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
-3.0
-2.9
-2.8
-2.7
-2.6
-2.5
-2.4
-2.3
-2.2
-2.1
-2.0
-1.9
-1.8
-1.7
-1.6
-1.5
-1.4
-1.3
-1.2
-1.1
-1.0
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
-4
-2
0
2
4
-3.0
-2.8
-2.6
-2.4
-2.2
-2.0
-1.8
-1.6
-1.4
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
y1
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
-3.0
-2.9
-2.8
-2.7
-2.6
-2.5
-2.4
-2.3
-2.2
-2.1
-2.0
-1.9
-1.8
-1.7
-1.6
-1.5
-1.4
-1.3
-1.2
-1.1
-1.0
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
-4
-2
0
2
4
-3.0
-2.8
-2.6
-2.4
-2.2
-2.0
-1.8
-1.6
-1.4
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
There is a left limit of $-1$, a right limit of $1$ but no limit at $c=0$.
$$~
\lim_{x \rightarrow 0} \frac{(3+x)^{-1} - (3-x)^{-1}}{x}
~$$
f(x) = ( (3 + x)^(-1) - (3-x)^(-1) ) / x
plot(f, -1, 1)
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
-3.0
-2.9
-2.8
-2.7
-2.6
-2.5
-2.4
-2.3
-2.2
-2.1
-2.0
-1.9
-1.8
-1.7
-1.6
-1.5
-1.4
-1.3
-1.2
-1.1
-1.0
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
-4
-2
0
2
4
-3.0
-2.8
-2.6
-2.4
-2.2
-2.0
-1.8
-1.6
-1.4
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
y1
-0.29
-0.28
-0.27
-0.26
-0.25
-0.24
-0.23
-0.22
-0.21
-0.20
-0.19
-0.18
-0.281
-0.280
-0.279
-0.278
-0.277
-0.276
-0.275
-0.274
-0.273
-0.272
-0.271
-0.270
-0.269
-0.268
-0.267
-0.266
-0.265
-0.264
-0.263
-0.262
-0.261
-0.260
-0.259
-0.258
-0.257
-0.256
-0.255
-0.254
-0.253
-0.252
-0.251
-0.250
-0.249
-0.248
-0.247
-0.246
-0.245
-0.244
-0.243
-0.242
-0.241
-0.240
-0.239
-0.238
-0.237
-0.236
-0.235
-0.234
-0.233
-0.232
-0.231
-0.230
-0.229
-0.228
-0.227
-0.226
-0.225
-0.224
-0.223
-0.222
-0.221
-0.220
-0.219
-0.218
-0.217
-0.216
-0.215
-0.214
-0.213
-0.212
-0.211
-0.210
-0.209
-0.208
-0.207
-0.206
-0.205
-0.204
-0.203
-0.202
-0.201
-0.200
-0.199
-0.198
-0.197
-0.196
-0.195
-0.194
-0.193
-0.192
-0.191
-0.190
-0.30
-0.25
-0.20
-0.15
-0.282
-0.280
-0.278
-0.276
-0.274
-0.272
-0.270
-0.268
-0.266
-0.264
-0.262
-0.260
-0.258
-0.256
-0.254
-0.252
-0.250
-0.248
-0.246
-0.244
-0.242
-0.240
-0.238
-0.236
-0.234
-0.232
-0.230
-0.228
-0.226
-0.224
-0.222
-0.220
-0.218
-0.216
-0.214
-0.212
-0.210
-0.208
-0.206
-0.204
-0.202
-0.200
-0.198
-0.196
-0.194
-0.192
-0.190
The answer is the $y$ value as $x$ goes to $0$, or by mousing about $-0.222$.
$$~
\lim_{x \rightarrow 0+} \log(\frac{1}{|x|})^{-1/32}
~$$
Here we need to be careful when we plot– we need a right limit and $f(1)$ is undefined, so we choose $(0,1/2)$:
f(x) = log(1/abs(x))^(-1/32)
plot(f, 0, 1/2)
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
-0.50
-0.48
-0.46
-0.44
-0.42
-0.40
-0.38
-0.36
-0.34
-0.32
-0.30
-0.28
-0.26
-0.24
-0.22
-0.20
-0.18
-0.16
-0.14
-0.12
-0.10
-0.08
-0.06
-0.04
-0.02
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0.18
0.20
0.22
0.24
0.26
0.28
0.30
0.32
0.34
0.36
0.38
0.40
0.42
0.44
0.46
0.48
0.50
0.52
0.54
0.56
0.58
0.60
0.62
0.64
0.66
0.68
0.70
0.72
0.74
0.76
0.78
0.80
0.82
0.84
0.86
0.88
0.90
0.92
0.94
0.96
0.98
1.00
-0.5
0.0
0.5
1.0
-0.50
-0.45
-0.40
-0.35
-0.30
-0.25
-0.20
-0.15
-0.10
-0.05
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
1.00
y1
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
-1.50
-1.45
-1.40
-1.35
-1.30
-1.25
-1.20
-1.15
-1.10
-1.05
-1.00
-0.95
-0.90
-0.85
-0.80
-0.75
-0.70
-0.65
-0.60
-0.55
-0.50
-0.45
-0.40
-0.35
-0.30
-0.25
-0.20
-0.15
-0.10
-0.05
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
1.55
1.60
1.65
1.70
1.75
1.80
1.85
1.90
1.95
2.00
2.05
2.10
2.15
2.20
2.25
2.30
2.35
2.40
2.45
2.50
2.55
2.60
2.65
2.70
2.75
2.80
2.85
2.90
2.95
3.00
-2
0
2
4
-1.5
-1.4
-1.3
-1.2
-1.1
-1.0
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
The limit looks to be about $0.94$ – but something looks amiss.
Using a table, identify the following limits. (You can also just look at $f(0.1), f(0.01), f(0.001), ...$)
$$~
\lim_{x \rightarrow 0} \frac{\sqrt{x + 25} - 5}{x}
~$$
We can make a table with lim, but that isn't a standard function. Rather, we just take some values near c:
f(x) = (sqrt(x + 25) - 5) / x
f(0.1), f(0.01), f(0.001), f(0.0001)
(0.09990019950139484,0.09999000199947261,0.0999990000201123,0.09999989999620595)
Looks like $0.1$ is the answer.
$$~
\lim_{x \rightarrow 0} \frac{|x|}{x}
~$$
If we aren't careful to look at both left and right we will be misled:
f(x) = abs(x)/x
f(0.1), f(0.01), f(0.001), f(0.0001)
(1.0,1.0,1.0,1.0)
shows 1.0 as the obvious answer, but this is to the question of a right limit. To see the left, we have:
f(-0.1), f(-0.01), f(-0.001), f(-0.0001)
(-1.0,-1.0,-1.0,-1.0)
with $-1.0$ as the answer. Two limits means no limit.
$$~
\lim_{x \rightarrow 0} \frac{(3+x)^{-1} - (3-x)^{-1}}{x}
~$$
f(x) = ( (3 + x)^(-1) - (3-x)^(-1) ) / x
f(0.1), f(0.01), f(0.001), f(0.0001)
(-0.2224694104560626,-0.22222469138545864,-0.2222222469135393,-0.22222222246959245)
Again we see in support of our graph, an answer of $-0.222... = -2/9$.
$$~
\lim_{x \rightarrow 0+} \log(\frac{1}{|x|})^{-1/32}
~$$
For the right limit we look close to $0$ on the right:
f(x) = log(1/abs(x))^(-1/32)
f(0.1), f(0.01), f(0.001), f(0.0001)
(0.9742732077388627,0.9533965419338172,0.9413924707884264,0.9329672200274579)
You might guess the answer is $0.93...$ from this.
Using SymPy find the following limits:
$$~
\lim_{x \rightarrow 0}\frac{\sqrt{1+x} - \sqrt{1-x}}{x}.
~$$
f(x) = (sqrt(1+x) - sqrt(1-x)) / x
limit(f, 0)
$$1$$
$$~
\lim_{x \rightarrow 4} \frac{x^{3/2} - 8}{x-4}
~$$
f(x) = (x^(3//2) - 8) / (x-4) ## this one is subtle x^(3/2) doens't work
limit(f, 4)
$$3$$
$$~
\lim_{x \rightarrow 0} \frac{|x|}{x}.
~$$
A nieve answer is wrong:
f(x) = abs(x)/x
limit(f, 0)
$$1$$
Why? The limit function gives the right limit by default. Here, the left limit isn't the same:
limit(f, 0, dir="-")
$$-1$$
So, no limit, as there are different left and right limits
$$~
\lim_{x \rightarrow 0+} \log(\frac{1}{|x|})^{-1/32}
~$$
f(x) = log(1/abs(x))^(-1/32)
limit(f, 0)
$$0$$
Wait, the graph and table showed the limit to be around $0.93$. Which is right?
Derivatives We assume you have run the following code:
using Roots
Base.ctranspose(f::Function) = D(f)
ctranspose (generic function with 40 methods)
Find the slope of the secant line of the function $f(x) = \sin(x)$ between $-\pi/3$ and $\pi/3$.
f(x) = sin(x)
a, b = -pi/3, pi/3
(f(b) - f(a)) / (b-a)
0.8269933431326881
Find the slope of the tangent line of the function $f(x) = \sin(x)$ at both $-\pi/3$ and $\pi/3$
f'(-pi/3), f'(pi/3)
(0.5000000000000001,0.5000000000000001)
Let $h=10^{-6}$. Find the approximate forward derivative of $f(x) = \sin(x)$ at $\pi/3$.
c = pi/3
h = 1e-6
(f(c+h) - f(c)) / h
0.4999995669718871
Close to the answer, but not precisely so.
What is the difference in value between your last number and that given by sin'(pi/3)?
fpapprox = (f(c+h) - f(c)) / h
f'(c) - fpapprox
4.330281130338065e-7
So the error is about the same size as the step size h.
An arrow is launched. The height as a function of time is given by:
$$~
y(t) = -16t^2 + 25t + 5, \quad t \geq 0
~$$
Find the rate of change of height at time $t=1$. Find the rate of change of height at the instant the arrow strikes the grount.
f(t) = -16t^2 + 25t + 5
f'(1)
-7.0
For the other, we need to first solve for $t$. We can see that between 0 (when the arrow is at 5) and $3$ (where $f(3) < 0$) we have a zero.
plot(f, 0, 3)
-4
-3
-2
-1
0
1
2
3
4
5
6
7
-3.0
-2.9
-2.8
-2.7
-2.6
-2.5
-2.4
-2.3
-2.2
-2.1
-2.0
-1.9
-1.8
-1.7
-1.6
-1.5
-1.4
-1.3
-1.2
-1.1
-1.0
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4.0
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5.0
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
6.0
-3
0
3
6
-3.0
-2.8
-2.6
-2.4
-2.2
-2.0
-1.8
-1.6
-1.4
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
3.6
3.8
4.0
4.2
4.4
4.6
4.8
5.0
5.2
5.4
5.6
5.8
6.0
y1
-200
-175
-150
-125
-100
-75
-50
-25
0
25
50
75
100
125
150
-175
-170
-165
-160
-155
-150
-145
-140
-135
-130
-125
-120
-115
-110
-105
-100
-95
-90
-85
-80
-75
-70
-65
-60
-55
-50
-45
-40
-35
-30
-25
-20
-15
-10
-5
0
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
105
110
115
120
125
-200
-100
0
100
200
-180
-170
-160
-150
-140
-130
-120
-110
-100
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
10
20
30
40
50
60
70
80
90
100
110
120
130
So $[0,3]$ is a bracketing interval:
t0 = fzero(f, 0, 3)
1.7419016343087124
We use this value to find the rate of change at:
f'(t0)
-30.740852297878796
Let $V(t) = 10(1 - t/100)^2$. Find the averate rate of change between $t=0$ and $t=100$. Call this value $a$. Using fzero find when $V'(t) = a$ between $0$ and $100$.
V(t) = 10 * (1 - t/100)^2
V (generic function with 1 method)
The average rate of change is just a secant line computation:
a, b= 0, 100
avg = (V(b) - V(a)) / (b-a)
-0.1
We use fzeros to find all values where $V(t)$ takes this value. For that we need a new function:
f(t) = V'(t) - avg
fzeros(f, a, b)
1-element Array{Float64,1}:
50.0
(The mean value theorem guarantees that we will find at least 1 zero here.) We graph to verify:
tangent(f,c) = x -> f(c) + f'(c) * (x-c)
secant(f, a, b) = x -> f(b) + (f(b)-f(a)) / (b-a) * (x-b)
plot(V, a, b, linewidth=5)
plot!(tangent(V, 50))
plot!(secant(V, 0, 100))
-150
-100
-50
0
50
100
150
200
250
-100
-95
-90
-85
-80
-75
-70
-65
-60
-55
-50
-45
-40
-35
-30
-25
-20
-15
-10
-5
0
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
105
110
115
120
125
130
135
140
145
150
155
160
165
170
175
180
185
190
195
200
-100
0
100
200
-100
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
y1
y2
y3
-25
-20
-15
-10
-5
0
5
10
15
20
25
30
-20.0
-19.5
-19.0
-18.5
-18.0
-17.5
-17.0
-16.5
-16.0
-15.5
-15.0
-14.5
-14.0
-13.5
-13.0
-12.5
-12.0
-11.5
-11.0
-10.5
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
12.5
13.0
13.5
14.0
14.5
15.0
15.5
16.0
16.5
17.0
17.5
18.0
18.5
19.0
19.5
20.0
20.5
21.0
21.5
22.0
22.5
23.0
23.5
24.0
24.5
25.0
-20
0
20
40
-20
-19
-18
-17
-16
-15
-14
-13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
First and second derivatives
Plot the function f(x) = x * airy(x) over the interval $[-3, 3]$. From your graph, answer: on what intervals is $f(x)$ positive ? increasing ? concave up ?
We plot and look:
f(x) = x * airy(x)
plot(f, -3, 3)
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
-9.0
-8.8
-8.6
-8.4
-8.2
-8.0
-7.8
-7.6
-7.4
-7.2
-7.0
-6.8
-6.6
-6.4
-6.2
-6.0
-5.8
-5.6
-5.4
-5.2
-5.0
-4.8
-4.6
-4.4
-4.2
-4.0
-3.8
-3.6
-3.4
-3.2
-3.0
-2.8
-2.6
-2.4
-2.2
-2.0
-1.8
-1.6
-1.4
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
3.6
3.8
4.0
4.2
4.4
4.6
4.8
5.0
5.2
5.4
5.6
5.8
6.0
6.2
6.4
6.6
6.8
7.0
7.2
7.4
7.6
7.8
8.0
8.2
8.4
8.6
8.8
9.0
-10
-5
0
5
10
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
y1
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
-3.5
-3.4
-3.3
-3.2
-3.1
-3.0
-2.9
-2.8
-2.7
-2.6
-2.5
-2.4
-2.3
-2.2
-2.1
-2.0
-1.9
-1.8
-1.7
-1.6
-1.5
-1.4
-1.3
-1.2
-1.1
-1.0
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4.0
-4
-2
0
2
4
-3.6
-3.4
-3.2
-3.0
-2.8
-2.6
-2.4
-2.2
-2.0
-1.8
-1.6
-1.4
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
3.6
3.8
4.0
positive: $(-3, -2.3)$ and $(0,3)$.
increasing: $(-1.5, 0.6)$
concave up: $(-2.4, -0.5)$ and $(1.7, 3)$.
These values are "eye-balled". You can get more precise if you used plotif: (plotif(f,f,a,b) for positive, plotif(f,f',a,b) for increasing and plotif(f,f'',a,b) for concave up.)
Find the critical point(s) of $f(x) = 2x - x^2/6 - x^3/9$. What are the values of $f''(x)$ at these point(s)?
The critical points are the zeros of $f'(x)$
f(x) = 2x - x^2/6 - x^3/9
plot(f, -4,4)
-14
-12
-10
-8
-6
-4
-2
0
2
4
6
8
10
12
14
-12.0
-11.5
-11.0
-10.5
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
-20
-10
0
10
20
-12.0
-11.5
-11.0
-10.5
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
y1
-15.0
-12.5
-10.0
-7.5
-5.0
-2.5
0.0
2.5
5.0
7.5
10.0
12.5
-12.5
-12.0
-11.5
-11.0
-10.5
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
-20
-10
0
10
-12.5
-12.0
-11.5
-11.0
-10.5
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
Plotting $f$ over the $(-4,4)$ shows the cubic polynomial will have both its peaks in that interval, so we can use fzeros there:
cps = fzeros(f', -4, 4)
2-element Array{Float64,1}:
-3.0
2.0
The values of the second derivative are:
map(f'', cps)
2-element Array{Float64,1}:
1.66667
-1.66667
The first is positive – relative minimum, the second negative – relative maximum – as the graph showed already.
Find the critical point(s) of $f(x) = 1/x + x, \quad x > 0$. What are the values of $f''(x)$ at these point(s)?
We plot to the right of $0$ – to avoid the asymptote – to look:
f(x) = 1/x + x
plot(f, .1, 3)
-4
-3
-2
-1
0
1
2
3
4
5
6
7
-3.0
-2.9
-2.8
-2.7
-2.6
-2.5
-2.4
-2.3
-2.2
-2.1
-2.0
-1.9
-1.8
-1.7
-1.6
-1.5
-1.4
-1.3
-1.2
-1.1
-1.0
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4.0
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5.0
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
6.0
-3
0
3
6
-3.0
-2.8
-2.6
-2.4
-2.2
-2.0
-1.8
-1.6
-1.4
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
3.6
3.8
4.0
4.2
4.4
4.6
4.8
5.0
5.2
5.4
5.6
5.8
6.0
y1
-20
-15
-10
-5
0
5
10
15
20
25
30
35
-15.0
-14.5
-14.0
-13.5
-13.0
-12.5
-12.0
-11.5
-11.0
-10.5
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
12.5
13.0
13.5
14.0
14.5
15.0
15.5
16.0
16.5
17.0
17.5
18.0
18.5
19.0
19.5
20.0
20.5
21.0
21.5
22.0
22.5
23.0
23.5
24.0
24.5
25.0
25.5
26.0
26.5
27.0
27.5
28.0
28.5
29.0
29.5
30.0
-20
0
20
40
-15
-14
-13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Looks like just one critical point and it is in this interval:
fzeros(f', .1, 3)
1-element Array{Float64,1}:
1.0
We have $f''(1) > 0$, so as the graph shows a relative minimum occurs at this critical point.
f''(1)
2.0
Find any values where the second derivative of $f(x) = 2x - x^2/6 - x^3/9$ is $0$.
f(x) = 2x - x^2/6 - x^3/9
plot(f, -6, 5)
-30
-25
-20
-15
-10
-5
0
5
10
15
20
25
-25.0
-24.5
-24.0
-23.5
-23.0
-22.5
-22.0
-21.5
-21.0
-20.5
-20.0
-19.5
-19.0
-18.5
-18.0
-17.5
-17.0
-16.5
-16.0
-15.5
-15.0
-14.5
-14.0
-13.5
-13.0
-12.5
-12.0
-11.5
-11.0
-10.5
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
12.5
13.0
13.5
14.0
14.5
15.0
15.5
16.0
16.5
17.0
17.5
18.0
18.5
19.0
19.5
20.0
-40
-20
0
20
-25
-24
-23
-22
-21
-20
-19
-18
-17
-16
-15
-14
-13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
y1
-35
-30
-25
-20
-15
-10
-5
0
5
10
15
20
25
30
35
-30
-29
-28
-27
-26
-25
-24
-23
-22
-21
-20
-19
-18
-17
-16
-15
-14
-13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
-40
-20
0
20
40
-30
-28
-26
-24
-22
-20
-18
-16
-14
-12
-10
-8
-6
-4
-2
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
The graph shows the "action" all happens between $-6$ and $5$, so we look there:
ips = fzeros(f'', -6, 5)
1-element Array{Float64,1}:
-0.5
The value of $-1/2$ is found. Clearly this will be an inflection point, as the graph changes concavity. If you don't believe this, we can graph to see $f''(x)$ changing sign at $-1/2$:
plot(f'', -6, 5)
scatter!(ips, 0*ips) # add the point
-30
-25
-20
-15
-10
-5
0
5
10
15
20
25
-25.0
-24.5
-24.0
-23.5
-23.0
-22.5
-22.0
-21.5
-21.0
-20.5
-20.0
-19.5
-19.0
-18.5
-18.0
-17.5
-17.0
-16.5
-16.0
-15.5
-15.0
-14.5
-14.0
-13.5
-13.0
-12.5
-12.0
-11.5
-11.0
-10.5
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
12.5
13.0
13.5
14.0
14.5
15.0
15.5
16.0
16.5
17.0
17.5
18.0
18.5
19.0
19.5
20.0
-40
-20
0
20
-25
-24
-23
-22
-21
-20
-19
-18
-17
-16
-15
-14
-13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
y1
y2
-14
-12
-10
-8
-6
-4
-2
0
2
4
6
8
10
12
14
-12.0
-11.5
-11.0
-10.5
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
-20
-10
0
10
20
-12.0
-11.5
-11.0
-10.5
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
Let $f(x) = x^3 - 3x + 5$. Find the critical points. Use the first derivative test to classify them as relative maxima, relative minima, or neither.
f(x) = x^3 - 3x + 5
cps = fzeros(f', -5, 5)
2-element Array{Float64,1}:
-1.0
1.0
We plot $f'$ to see how it changes sign:
plot(f', -2, 2)
scatter!(cps, 0*cps) # add points
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
-6.0
-5.8
-5.6
-5.4
-5.2
-5.0
-4.8
-4.6
-4.4
-4.2
-4.0
-3.8
-3.6
-3.4
-3.2
-3.0
-2.8
-2.6
-2.4
-2.2
-2.0
-1.8
-1.6
-1.4
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
3.6
3.8
4.0
4.2
4.4
4.6
4.8
5.0
5.2
5.4
5.6
5.8
6.0
-6
-3
0
3
6
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
y1
y2
-25
-20
-15
-10
-5
0
5
10
15
20
25
30
-20.0
-19.5
-19.0
-18.5
-18.0
-17.5
-17.0
-16.5
-16.0
-15.5
-15.0
-14.5
-14.0
-13.5
-13.0
-12.5
-12.0
-11.5
-11.0
-10.5
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
12.5
13.0
13.5
14.0
14.5
15.0
15.5
16.0
16.5
17.0
17.5
18.0
18.5
19.0
19.5
20.0
20.5
21.0
21.5
22.0
22.5
23.0
23.5
24.0
24.5
25.0
-20
0
20
40
-20
-19
-18
-17
-16
-15
-14
-13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
At $-1$ the derivative changes sign from plus to minus – a relative max for $f(x)$.
At $1$ the derivative changes sign from minus to plus – a relative min.
Let $f(x) = x^2 e^{-x/3}$. Find the critical points. Use the first derivative test to classify them as relative maxima, relative minima, or neither.
A graph over $[-3, 20]$ shows the general shape. (Picked by trial and error!)
f(x) = x^2 * exp(-x/3)
plot(f, -3, 20)
-35
-30
-25
-20
-15
-10
-5
0
5
10
15
20
25
30
35
40
45
50
-30
-29
-28
-27
-26
-25
-24
-23
-22
-21
-20
-19
-18
-17
-16
-15
-14
-13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
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30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
-50
0
50
-30
-28
-26
-24
-22
-20
-18
-16
-14
-12
-10
-8
-6
-4
-2
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
y1
-30
-25
-20
-15
-10
-5
0
5
10
15
20
25
30
35
40
45
50
55
-25
-24
-23
-22
-21
-20
-19
-18
-17
-16
-15
-14
-13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
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20
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30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
-25
0
25
50
-26
-24
-22
-20
-18
-16
-14
-12
-10
-8
-6
-4
-2
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
There should be two critical points:
cps = fzeros(f', -3, 20)
2-element Array{Float64,1}:
-0.0
6.0
The graph of the derivative shows the change of sign:
plot(f', -1, 7)
scatter!(cps, 0*cps)
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
12.5
13.0
13.5
14.0
14.5
15.0
-10
0
10
20
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
12.5
13.0
13.5
14.0
14.5
15.0
y1
y2
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
-10.0
-9.8
-9.6
-9.4
-9.2
-9.0
-8.8
-8.6
-8.4
-8.2
-8.0
-7.8
-7.6
-7.4
-7.2
-7.0
-6.8
-6.6
-6.4
-6.2
-6.0
-5.8
-5.6
-5.4
-5.2
-5.0
-4.8
-4.6
-4.4
-4.2
-4.0
-3.8
-3.6
-3.4
-3.2
-3.0
-2.8
-2.6
-2.4
-2.2
-2.0
-1.8
-1.6
-1.4
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
3.6
3.8
4.0
4.2
4.4
4.6
4.8
5.0
5.2
5.4
5.6
5.8
6.0
6.2
6.4
6.6
6.8
7.0
7.2
7.4
7.6
7.8
8.0
-10
-5
0
5
10
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
The critical point of 0 sees a change of sign in $f'(x)$ from negative to positive, so is a relative min for $f$.
The critical point of 6 sees a change of sign in $f'(x)$ from positive to negative, so is a relative max for $f$.
Here the second derivative test is easier:
map(f'', cps) # plus -> relative min; minus -> relative max
2-element Array{Float64,1}:
2.0
-0.270671
For the function f(x) = (x^2)^(1/3)*(2-x). Make a plot over $[-4,4]$ and identify graphically any critical points. Are all the critical points found by fzeros(f, -4, 4)? Can the second derivative test classify all the critical points? Can the first derivative test classify all the critical points?
f(x) = (x^2)^(1/3)*(2-x)
plot(f, -4, 4)
-14
-12
-10
-8
-6
-4
-2
0
2
4
6
8
10
12
14
-12.0
-11.5
-11.0
-10.5
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
-20
-10
0
10
20
-12.0
-11.5
-11.0
-10.5
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
y1
-50
-40
-30
-20
-10
0
10
20
30
40
50
60
-40
-39
-38
-37
-36
-35
-34
-33
-32
-31
-30
-29
-28
-27
-26
-25
-24
-23
-22
-21
-20
-19
-18
-17
-16
-15
-14
-13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
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33
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37
38
39
40
41
42
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44
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46
47
48
49
50
-50
0
50
-40
-38
-36
-34
-32
-30
-28
-26
-24
-22
-20
-18
-16
-14
-12
-10
-8
-6
-4
-2
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
At $x=0$ this function does not have a derivative, so $0$ is a critical point – but it is not a zero of $f'(x)$. Will it still be found by fzeros?
cps = fzeros(f', -4, 4)
2-element Array{Float64,1}:
0.0
0.8
Yup. We can't use the second derivative test:
map(f'', cps)
2-element Array{Float64,1}:
NaN
-1.79536
That NaN should be expected – the first derivative isn't defined there, so the second can't be either. But we can look at the change of sign in the first derivative and apply the first derivative test:
plot(f', -4, 4)
scatter!(cps, 0*cps)
-14
-12
-10
-8
-6
-4
-2
0
2
4
6
8
10
12
14
-12.0
-11.5
-11.0
-10.5
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
-20
-10
0
10
20
-12.0
-11.5
-11.0
-10.5
-10.0
-9.5
-9.0
-8.5
-8.0
-7.5
-7.0
-6.5
-6.0
-5.5
-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
y1
y2
-35
-30
-25
-20
-15
-10
-5
0
5
10
15
20
25
30
35
-30
-29
-28
-27
-26
-25
-24
-23
-22
-21
-20
-19
-18
-17
-16
-15
-14
-13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
-40
-20
0
20
40
-30
-28
-26
-24
-22
-20
-18
-16
-14
-12
-10
-8
-6
-4
-2
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
The change of sign at $0$ is obvious if you ignore the spurious vertical line that is graphed. The other change of sign from positive to negative indicates a relative max, as can be seen in the original graph of $f(x)$.