Solutions to Assignment #08 –MATH 1401
Kawai
Section 3.3
(#2) f (x) =
x2
:
(x 1)2
We have three interesting intervals over which we must …nd the absolute minimum and
maximum points. Here are the correct answers!
Minimum Point?
(x; y)
Interval
Maximum Point?
(x; y)
( 1; +1)
(0; 0)
[ 1; 1]
(0; 0)
(0; 1)
Tricky!
d:n:e:
Interval does not include
(0; 0) :
d:n:e:
Vertical asymptote at
x = 1:
d:n:e:
Vertical asymptote at
x = 1:
d:n:e:
Vertical asymptote at
x = 1:
Since f has a vertical asymptote at x = 1; we see that
lim f (x) = lim f (x) = +1:
x!1+
x!1
Thus, any interval which contains the points on either side of x = 1 cannot have an absolute
maximum point.
Let’s show that x = 0 is a critical number.
f 0 (x) =
=
=
(x
0
x2
2x (x
2x2
1)2 x2 (2 (x
(x 1)4
2x2
3
1)
=
1))
h
x2 (x
1)2
(x
2x
(x
0
=
1)2
(x
1)2 x2
=
(x
1)2
2
i0
1) 2x (x
(x
1)
2x2
1)4
2x
= 0:
(x 1)3
The critical number is x = 0:X
Even though we hadn’t seen this part yet in lecture, let’s consider applying the First Derivative
Test.
We should be able to sketch
f 0 (x) =
1
2x
:
(x 1)3
Since there is a vertical asymptote at x = 1; we have one critical point and one point of
interest. This divides the number line into three intervals.
( 1; 0)
( )
&
f 0 (x)
(0; 1)
(+)
%
(1; +1)
( )
&
Since f is continuous everywhere except at x = 1; we must have a local minimum at
x = 0: The function decreases going in and increases coming out! [Remember that the
tangent line is horizontal when x = 0:]
Also, note that we can …nd the horizontal asymptotes for f: The degree of the numerator is
1 and the degree of the denominator is 3. Since it is “bottom heavy”, we can apply L’Hôpital
multiple times and the numerator would eventually become zero. Thus, we have the left and
right end behaviors:
lim f (x) = lim f (x) = 0:
x! 1
x!+1
(#12) Find all critical numbers by hand (since you can factor, correct?)
Annotate appropriate points of interest on your graph.
f (x) = x4 + 6x2
0
2
3
f (x) = 4x + 12x = 4x x2 + 3 = 0:
Since f is a polynomial, we know that it is continuous and di¤erentiable everywhere.
The last equation gives us x = 0 as its only critical number since the other factor would give
us nonreal roots.
Here’s the chart...
( 1; 0)
( )
&
f 0 (x)
(0; +1)
(+)
%
Since x2 + 3 is always positive, the sign of f 0 (x) only depends on x:
We see that x = 0 must be a local minimum point, (0;
y
4
2
0
-4
2) :
-2
0
2
4
x
-2
-4
We note that since there are no other critical numbers,
there are no other chances for the graph of f to change
directions.
Thus, we can also say that its concavity must always
be positive (concave up).
Later, we will see that this is equivalent to saying that
f 00 (x) = 12x2 + 12 is always positive.
Thus, the point (0;
2
2) is also the absolute minimum.
(#14) f (x) = x2=5
3x1=5
2
:
We should notice a few things immediately.
Since these are …fth roots, the domain is all the real numbers, ( 1; +1) or R:
Since this is a perfect square, the absolute minimum is probably zero. By inspection, we see
that f (0) = 0 so we know that f has at least one absolute minimum at (0; 0) : It might have
more than one.
Find the critical numbers.
f 0 (x) = 2 x2=5
3x1=5
h
x2=5
3x1=5
i0
= 2 x2=5
2
x
5
3x1=5
3=5
3
x
5
4=5
= 0:
This gives us two equations to solve. We set each factor equal to zero.
x2=5
3x1=5 = 0
x1=5 x1=5
3
= 0
Typically, we factor out the smallest positive exponent when all exponents are positive.
This gives us
x1=5 = 0 ) x = 0
and
x1=5
3 = 0 ) x1=5 = 3 ) x = 35 = 243:
The second factor gives us
2
x
5
3=5
3
x
5
4=5
=0)
2
5x3=5
3
=0
5x4=5
If we multiply through by what would be the LCD, 5x4=5 ; then we can eliminate all of the
denominators in one turn.
2 5x4=5
5x3=5
3 5x4=5
5x4=5
= 0 5x4=5 ) 2x1=5
x1=5 =
3
)x=
2
3
2
5
=
3=0
243
:
32
Thus we should have three critical numbers.
y
6
We have a sharp turn/vertical tangent line at x = 0:
4
We have a local maximum at x =
2
0
-5
0
5
10
15
20
243 :
= 7:6:
32
We must look much farther to the right if we want to see
the last critical number.
x
-2
3
y 0.002
From the scaling on the y-axis, we see that the curve is
always fairly close to the x-axis up through x = 243:
We achieve an absolute minimum at (243; 0) :
0.0015
0.001
We note that between the local maximum and local
minimum points, there must be an in‡ection point.
0.0005
0
240
242
244
246
248
250
x
(#20) Quotient Rule.
f (x) =
0
f (x) =
x2
x+4
x 1
(x
1) x2
x+4
0
(x
1) (2x
2x2
=
x + 4 [x
1]0
1)2
(x
=
x2
1)
x2
(x
2
x + 4 (1)
1)
3x + 1 x2 + x
(x 1)2
4
=
(x + 1) (x 3)
:
(x 1)2
By substitution, we already know that x = 1 is a vertical asymptote.
The factored numerator of f 0 (x) gives us x =
1; 3 as critical numbers.
Thus, we have two critical numbers and one point of interest.
( 1; 1)
(+)
%
f 0 (x)
y
( 1; 1)
( )
&
(3; +1)
(+)
%
8
7
6
5
4
3
2
1
0
-2
(1; 3)
( )
&
-1
0
-1
-2
-3
-4
1
2
3
4
When we are located to the left of x = 1; both numerator
factors are negative, and f 0 (x) is positive.
When we are located to the right of x = 3; both numerator
factors are positive, and f 0 (x) is positive.
When we are between x = 1 and x = 3; the numerator is
negative.
x
Thus, we must have a local maximum at ( 1;
local minimum at (3; 5) :
3) and a
-5
Again, we note that the chart correctly predicts the behavior of f on both sides of the
asymptote.
4
(#24) Product Rule. We see that f is continuous and di¤erentiable everywhere.
2x
f (x) = xe
f 0 (x) = x e
2x 0
+ [x]0 e
2x
= ( 2x + 1) e
Since e
2x
2x
=x
2x
2e
+e
2x
= 0:
is always positive, we can divide it out on both sides. This leaves
1
2x + 1 = 0 ) x = :
2
1;
1
; +1
2
1
2
(+)
%
f 0 (x)
y
( )
&
2
The chart correctly predicts a local maximum at
1
0
-2
-1
0
1
2
x
We note that we have a horizontal asymptote for the
right end behavior by L’Hôpital:
x
= 0:
x!+1 e2x
-1
lim
-2
(#25) f (x) = x4=3 + 4x1=3 + 4x
2=3 :
Immediately, we should recognize that
2=3
4x
=
4
x2=3
and we have a vertical asymptote at x = 0: So x = 0 is a point of interest.
Find the critical numbers.
f 0 (x) =
=
1 1
;
2 2e
4 1=3
x +4
3
1
x
3
2=3
4 1=3
4
x + 2=3
3
3x
+4
2
x
3
5=3
8
= 0:
3x5=3
We should multiply through by 3x5=3 :
8 3x5=3
3x5=3
4 3x5=3
4 1=3
x
3x5=3 +
3
3x2=3
4x2 + 4x
4 (x + 2) (x
5
= 0 3x5=3
8 = 0
1) = 0 ) x =
2; 1:
:
f 0 (x)
( 1; 2)
( )
&
y
( 2; 0)
(+)
%
(0; 1)
( )
&
(1; +1)
(+)
%
15
The chart correctly predicts the local minima at
( 2; 0) and (1; 9) :
10
We have a vertical asymptote at x = 0:
Both end behaviors approach (+1) :
5
0
-3
-2
-1
0
1
2
3
x
(#36) Absolute extrema.
f (x) = x4
f 0 (x) = 4x3
8x2 + 2
16x = 4x x2
The critical numbers are x =
2; 0; 2:
( 1; 2)
( )
&
(0; 2)
( )
&
f 0 (x)
y
-3
-2
-1
( 2; 0)
(+)
%
4 = 4x (x + 2) (x
2) = 0
(2; +1)
(+)
%
5
0
0
1
2
x
3
This will give us local minima at ( 2;
and a local maximum at (0; 2) :
14) and (2;
14)
-5
Now consider the absolute extrema situations below.
-10
-15
(a) The interval is [ 3; 1] : This contains the local minimum at ( 2; 14) and the local
maximum at (0; 2) :
According to the Extreme Value Theorem and Fermat, we need only check the values of
f at the critical points in the interval and at the end points.
f ( 3) = 11
f (1) =
5
The lowest value of f is designated at the absolute minimum. This occurs at ( 2; 14) :
The highest value of f is designated at the absolute maximum. This occurs at ( 3; 11) :
6
(b) The interval is [ 1; 3] : This contains the local maximum at (0; 2) and the local minimum at (2; 14) :
We check the end points.
f ( 1) =
5
f (3) = 11
The lowest value of f is designated at the absolute minimum. This occurs at (2; 14) :
The highest value of f is designated at the absolute maximum. This occurs at (3; 11) :
(#68) A measure of unpredictability is called entropy. If 0 < x < 1; then we have
H (x) =
x ln (x)
(1
x) ln (1
x) :
If x corresponds to the probability that there are NO transmission/reception errors when …les
are downloaded over the Internet, then H (x) gives us a measure of unpredictability for the
system.
We note that even though H (x) is not de…ned for x = 0 and x = 1; we have
lim x ln (x) = lim
x!0+
x!0+
ln (x)
:
1
x
By L’Hôpital, this gives us
lim
x!0+
1
x
1
x2
ln (x)
= lim
1
x!0+
x
= lim ( x) = 0:
x!0+
Similarly,
lim (1
x!1
x) ln (1
x) = 0:
The entropy graph looks like this:
y
1
0 .9
0 .8
If the system were “perfect”, then x = 1 and there
would be no unpredictability.
From the graph, we see that the maximum entropy
occurs when x = 0:5:
We want to show this algebraically using the
derivative.
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
x
7
Find the critical number.
H 0 (x) =
[x ln (x)]0
[(1
x) ln (1
=
x [ln (x)]0 + [x]0 ln (x)
=
x
1
x
(1
+ ln (x)
=
(1 + ln (x))
=
ln (x) + ln (1
x)]0
(1
( 1
x) [ln (1
[1
1
x)
ln (1
x]0
x
x)]0 + [1
x]0 ln (1
+ ( 1) ln (1
x)
x))
x) = 0:
Using the logarithm rules, we have
ln
1
x
= 0:
x
We exponentiate both sides.
ln
e
1 x
x
= e0 )
1
x
x
1
= 1 ) x = :X
2
If we test H 0 (x) with x = 0:1 and x = 0:9; then we have the resulting chart.
f 0 (x)
(0; 0:5)
(+)
%
(0:5; 1:0)
( )
&
H (x) attains its maximum value at (0:5; ln (2)) :
8
x)