Relative Extrema Edit
Find the relative maximum(s) and minimum(s), if any, of the following functions.
1.
f
(
x
)
=
x
x
+
1
{\displaystyle f(x)={\frac {x}{x+1}}}
2.
f
(
x
)
=
(
x
−
1
)
2
3
{\displaystyle f(x)={\sqrt[{3}]{(x-1)^{2}}}}
Minimum at the point
(
1
,
0
)
{\displaystyle (1,0)}
3.
f
(
x
)
=
x
2
+
2
x
{\displaystyle f(x)=x^{2}+{\frac {2}{x}}}
Relative minimum at
x
=
1
{\displaystyle x=1}
4.
f
(
s
)
=
s
1
+
s
2
{\displaystyle f(s)={\frac {s}{1+s^{2}}}}
Relative minimum at
s
=
−
1
{\displaystyle s=-1}
s
=
1
{\displaystyle s=1}
5.
f
(
x
)
=
x
2
−
4
x
+
9
{\displaystyle f(x)=x^{2}-4x+9}
Relative minimum at
x
=
2
{\displaystyle x=2}
6.
f
(
x
)
=
x
2
+
x
+
1
x
2
−
x
+
1
{\displaystyle f(x)={\frac {x^{2}+x+1}{x^{2}-x+1}}}
Relative minimum at
x
=
−
1
{\displaystyle x=-1}
x
=
1
{\displaystyle x=1}
Solutions
Range of Function Edit
7. Show that the expression
x
+
1
x
{\displaystyle x+{\frac {1}{x}}}
cannot take on any value strictly between 2 and -2.
f
(
x
)
=
x
+
1
x
{\displaystyle f(x)=x+{\frac {1}{x}}}
f
′
(
x
)
=
1
−
1
x
2
{\displaystyle f'(x)=1-{\frac {1}{x^{2}}}}
1
−
1
x
2
=
0
⟹
x
=
±
1
{\displaystyle 1-{\frac {1}{x^{2}}}=0\implies x=\pm 1}
f
″
(
x
)
=
2
x
3
{\displaystyle f''(x)={\frac {2}{x^{3}}}}
f
″
(
−
1
)
=
−
2
{\displaystyle f''(-1)=-2}
f
″
(
−
1
)
{\displaystyle f''(-1)}
x
=
−
1
{\displaystyle x=-1}
f
(
−
1
)
=
−
2
{\displaystyle f(-1)=-2}
lim
x
→
−
∞
f
(
x
)
=
−
∞
{\displaystyle \lim \limits _{x\to -\infty }f(x)=-\infty }
x
<
−
1
{\displaystyle x<-1}
f
′
(
x
)
{\displaystyle f'(x)}
x
{\displaystyle x}
f
{\displaystyle f}
−
2
{\displaystyle -2}
x
=
−
1
{\displaystyle x=-1}
−
1
<
x
<
1
{\displaystyle -1<x<1}
f
′
(
x
)
{\displaystyle f'(x)}
lim
x
→
0
−
f
(
x
)
=
−
∞
{\displaystyle \lim _{x\to 0^{-}}f(x)=-\infty }
lim
x
→
0
+
f
(
x
)
=
∞
{\displaystyle \lim _{x\to 0^{+}}f(x)=\infty }
f
″
(
1
)
=
2
{\displaystyle f''(1)=2}
f
″
(
1
)
{\displaystyle f''(1)}
x
=
1
{\displaystyle x=1}
f
(
1
)
=
2
{\displaystyle f(1)=2}
[
−
1
,
0
)
{\displaystyle [-1,0)}
−
2
{\displaystyle -2}
−
∞
{\displaystyle -\infty }
+
∞
{\displaystyle +\infty }
2
{\displaystyle 2}
x
=
1
{\displaystyle x=1}
x
>
1
{\displaystyle x>1}
f
′
(
x
)
{\displaystyle f'(x)}
2
{\displaystyle 2}
−
2
{\displaystyle -2}
2
{\displaystyle 2}
Absolute Extrema Edit
Determine the absolute maximum and minimum of the following functions on the given domain
8.
f
(
x
)
=
x
3
3
−
x
2
2
+
1
{\displaystyle f(x)={\frac {x^{3}}{3}}-{\frac {x^{2}}{2}}+1}
on
[
0
,
3
]
{\displaystyle [0,3]}
Maximum at
(
3
,
11
2
)
{\displaystyle (3,{\tfrac {11}{2}})}
(
1
,
5
6
)
{\displaystyle (1,{\tfrac {5}{6}})}
9.
f
(
x
)
=
(
4
3
x
2
−
1
)
x
{\displaystyle f(x)=\left({\frac {4}{3}}x^{2}-1\right)x}
on
[
−
1
2
,
2
]
{\displaystyle [-{\tfrac {1}{2}},2]}
Maximum at
(
2
,
26
3
)
{\displaystyle (2,{\tfrac {26}{3}})}
(
1
2
,
−
1
3
)
{\displaystyle ({\tfrac {1}{2}},-{\tfrac {1}{3}})}
Solutions
Determine Intervals of Change Edit
Find the intervals where the following functions are increasing or decreasing
10.
f
(
x
)
=
10
−
6
x
−
2
x
2
{\displaystyle f(x)=10-6x-2x^{2}}
Increasing on
(
−
∞
,
−
3
2
)
{\displaystyle (-\infty ,-{\tfrac {3}{2}})}
(
−
3
2
,
∞
)
{\displaystyle (-{\tfrac {3}{2}},\infty )}
11.
f
(
x
)
=
2
x
3
−
12
x
2
+
18
x
+
15
{\displaystyle f(x)=2x^{3}-12x^{2}+18x+15}
Decreasing on
(
1
,
3
)
{\displaystyle (1,3)}
12.
f
(
x
)
=
5
+
36
x
+
3
x
2
−
2
x
3
{\displaystyle f(x)=5+36x+3x^{2}-2x^{3}}
Increasing on
(
−
2
,
3
)
{\displaystyle (-2,3)}
13.
f
(
x
)
=
8
+
36
x
+
3
x
2
−
2
x
3
{\displaystyle f(x)=8+36x+3x^{2}-2x^{3}}
Increasing on
(
−
2
,
3
)
{\displaystyle (-2,3)}
14.
f
(
x
)
=
5
x
3
−
15
x
2
−
120
x
+
3
{\displaystyle f(x)=5x^{3}-15x^{2}-120x+3}
Decreasing on
(
−
2
,
4
)
{\displaystyle (-2,4)}
15.
f
(
x
)
=
x
3
−
6
x
2
−
36
x
+
2
{\displaystyle f(x)=x^{3}-6x^{2}-36x+2}
Decreasing on
(
−
2
,
6
)
{\displaystyle (-2,6)}
Solutions
Determine Intervals of Concavity Edit
Find the intervals where the following functions are concave up or concave down
16.
f
(
x
)
=
10
−
6
x
−
2
x
2
{\displaystyle f(x)=10-6x-2x^{2}}
17.
f
(
x
)
=
2
x
3
−
12
x
2
+
18
x
+
15
{\displaystyle f(x)=2x^{3}-12x^{2}+18x+15}
Concave down on
(
−
∞
,
2
)
{\displaystyle (-\infty ,2)}
(
2
,
∞
)
{\displaystyle (2,\infty )}
18.
f
(
x
)
=
5
+
36
x
+
3
x
2
−
2
x
3
{\displaystyle f(x)=5+36x+3x^{2}-2x^{3}}
Concave up on
(
−
∞
,
1
2
)
{\displaystyle (-\infty ,{\tfrac {1}{2}})}
(
1
2
,
∞
)
{\displaystyle ({\tfrac {1}{2}},\infty )}
19.
f
(
x
)
=
8
+
36
x
+
3
x
2
−
2
x
3
{\displaystyle f(x)=8+36x+3x^{2}-2x^{3}}
Concave up on
(
−
∞
,
1
2
)
{\displaystyle (-\infty ,{\tfrac {1}{2}})}
(
1
2
,
∞
)
{\displaystyle ({\tfrac {1}{2}},\infty )}
20.
f
(
x
)
=
5
x
3
−
15
x
2
−
120
x
+
3
{\displaystyle f(x)=5x^{3}-15x^{2}-120x+3}
Concave down on
(
−
∞
,
1
)
{\displaystyle (-\infty ,1)}
(
1
,
∞
)
{\displaystyle (1,\infty )}
21.
f
(
x
)
=
x
3
−
6
x
2
−
36
x
+
2
{\displaystyle f(x)=x^{3}-6x^{2}-36x+2}
Concave down on
(
−
∞
,
2
)
{\displaystyle (-\infty ,2)}
(
2
,
∞
)
{\displaystyle (2,\infty )}
Solutions
Word Problems Edit
22. You peer around a corner. A velociraptor 64 meters away spots you. You run away at a speed of 6 meters per second. The raptor chases, running towards the corner you just left at a speed of
4
t
{\displaystyle 4t}
meters per second (time
t
{\displaystyle t}
measured in seconds after spotting). After you have run 4 seconds the raptor is 32 meters from the corner. At this time, how fast is death approaching your soon to be mangled flesh? That is, what is the rate of change in the distance between you and the raptor?
10
m
s
{\displaystyle 10{\tfrac {m}{s}}}
23. Two bicycles leave an intersection at the same time. One heads north going
12
m
p
h
{\displaystyle 12{\rm {mph}}}
and the other heads east going
5
m
p
h
{\displaystyle 5{\rm {mph}}}
. How fast are the bikes getting away from each other after one hour?
13
m
p
h
{\displaystyle 13{\rm {mph}}}
24. You're making a can of volume
200
m
3
{\displaystyle 200m^{3}}
with a gold side and silver top/bottom. Say gold costs 10 dollars per m
2
{\displaystyle ^{2}}
and silver costs 1 dollar per
m
2
{\displaystyle m^{2}}
. What's the minimum cost of such a can?
Solutions
Graphing Functions Edit
For each of the following, graph a function that abides by the provided characteristics
25.
f
(
1
)
=
f
(
−
2
)
=
0
,
lim
x
→
∞
f
(
x
)
=
lim
x
→
−
∞
f
(
x
)
=
0
,
vertical asymptote at
x
=
−
3
,
f
′
(
x
)
>
0
on
(
0
,
2
)
,
{\displaystyle f(1)=f(-2)=0,\ \lim _{x\to \infty }f(x)=\lim _{x\to -\infty }f(x)=0,\ {\mbox{ vertical asymptote at }}x=-3,\ f'(x)>0{\mbox{ on }}(0,2),}
f
′
(
x
)
<
0
on
(
−
∞
,
−
3
)
∪
(
−
3
,
0
)
∪
(
2
,
∞
)
,
f
″
(
x
)
>
0
on
(
−
3
,
1
)
∪
(
3
,
∞
)
,
f
″
(
x
)
<
0
on
(
−
∞
,
−
3
)
∪
(
1
,
3
)
.
{\displaystyle f'(x)<0{\mbox{ on }}(-\infty ,-3)\cup (-3,0)\cup (2,\infty ),\ f''(x)>0{\mbox{ on }}(-3,1)\cup (3,\infty ),\ f''(x)<0{\mbox{ on }}(-\infty ,-3)\cup (1,3).}
There are many functions that satisfy all the conditions. Here is one example:
26.
f
has domain
[
−
1
,
1
]
,
f
(
−
1
)
=
−
1
,
f
(
−
1
2
)
=
−
2
,
f
′
(
−
1
2
)
=
0
,
f
″
(
x
)
>
0
on
(
−
1
,
1
)
{\displaystyle f{\mbox{ has domain }}[-1,1],\ f(-1)=-1,\ f(-{\tfrac {1}{2}})=-2,\ f'(-{\tfrac {1}{2}})=0,\ f''(x)>0{\mbox{ on }}(-1,1)}
There are many functions that satisfy all the conditions. Here is one example:
Solutions
The 
![{\displaystyle f(x)={\sqrt[{3}]{(x-1)^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4371183376851214640d374c3917162fc7ee8ac)
![[0,3]](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5c9e70f7d437509d4ebedb0eaf7ada946e91a79)
![{\displaystyle [-{\tfrac {1}{2}},2]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad45dba0e31544218eea18079f13e69dc265793f)
![{\displaystyle f{\mbox{ has domain }}[-1,1],\ f(-1)=-1,\ f(-{\tfrac {1}{2}})=-2,\ f'(-{\tfrac {1}{2}})=0,\ f''(x)>0{\mbox{ on }}(-1,1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ef253d531e7a7682bcad1a7d37ccea25f15a98d)
