Download Math 141

Math 150
Must Show all Work!
Quiz
Name ____________________
1. Sketch the graph
x2  3
y=
x3
x-intercepts: (  3 ,0)
Vertical Asymptote: x = 0
Horizontal Asymptote: y = 0
[2 x]( x 3 )  ( x 2  3)[3x 2 ]
y’ =
=
(x3 )2
=
 x 4  9x 2
x6
9  x 2 (3  x)(3  x)


x4
x4
 the critical numbers are 3 and -3 
-----
++++
-
------
++++
-3
0
y’
3
 By the First Derivative Test, we have a local minimum at x = -3 and a local maximum at x = 3.
The local minimum value is -0.22 and the local maximum is 0.22.
 Local Minimum (-3, -0.22) and Local Maximum (3, 0.22) .
1 x
2. Describe the intervals on which f(x) =
1 x
is concave up and concave down. Also list the points
of inflection.
1
1
1
1
1
1
1
1
1 
1 
1 
1 
[ x 2 ](1  x 2 )  (1  x 2 )[ x 2 ] [ x 2 ](1  x 2 )  (1  x 2 )[ x 2 ]
2
2
2
2
f’(x) =
=
1
1
(1  x 2 ) 2
(1  x 2 ) 2
(Multiply numerator and denominator to clear the smaller fractions)
 f’(x) =
1
2
1
2
(1  x )  (1  x )
1
2
1
2
2 x (1  x ) 2
3
1
=
1
1
2
1
2
 x

1
2
1
2
* (1  x ) 
x (1  x ) 2
1
1
1
1 
1 
1 
 f” (x) = [ x 2 ](1+ x 2 )-2 +(- x 2 )[-2( 1  x 2 ) 3 * x 2 ] =
2
2
2
2
 12
 2x
 1
 2x 2






=

3
2
1
2
1
2

3
2
1
2
1
2
1
1
x (1+ x )-2 +x-1 ( 1  x ) 3 = x (1+ x )-3 [(1+ x ) + 2 x ] =
2
2
Function
undefined
+++++++++++
1  3x
1
2
3
1
2
1
2 x 2 (1  x 2 ) 3
Though not very exciting, we have
shown that this function is concave
up throughout its domain.
y’’
0
Here is the graph:
3. Describe the intervals on which f(x) = 2 cos 2 x  x 2 for x in [0,  ] is concave up and concave down.
Also list the points of inflection.
f’(x) = -4cosxsinx-2x = -2sin(2x) – 2x
f’' (x) = -4cos(2x) – 2
PPI: cos(2x) = -
 2

  =  3
4

 3
1
1

 by letting  = 2x  cos(  ) =   ’ =
2
2
3
 2k
 2k
 2x =
(***Correction above***)
 2
 3
 4

 3
 2k
 2k
 x =

 3  k
 2

 k
 3
 we get only 2 possible point of inflection x =  and 2 that lies with in the interval [0,  ]
3
3
 sign chart to determine concavity :
Concave down
------------
Concave up Conc. Down
-------+++++

3
2
3
We get two points of inflection

2
which is ( ,-0.60) & (
, -3.89)
3
3
y’’
4. A window has the shape of a rectangle surmounted by a semicircle. If the total perimeter of the
window is 20 feet, find the dimensions that will maximize the amount of light to enter.
Let x represent the radius of the semicircle and y represent the height of the rectangular portion
of the window:
The function to be maximized is the area A = 2xy +
1
 x2
2
Since the perimeter of the window is 20 feet we have 20 = 2y + 2x +  x
We can use this to rewrite y in terms of x: y = 10 – x -

x (*)
2
Substituting (*) into the area function we obtain:

1
1
x ) +  x2 = 20x -2x2 -  x2 +  x2
2
2
2
(Note that this is a quadratic function which means that the only critical number occurs at the
vertex)
A(x) = 2x(10 – x -
 A’(x) = 20 – 4x - 2  x +  x = 20 - 4x -  x
 by setting this equal to zero we obtain the critical number x =
20
4
 2.80
 since A’’ is easy to calculate, let’s use the Second Derivative Test to verify that this critical
number results in a Maximum Area : A’’(x) = -4 -  which shows that this function is always
concave down and that the Local Maximum and Absolute Maximum Area occur at
20
 2.80
x=
4
20
 2.80
To find y we substitute this x value into (*) to obtain y = x =
4
Therefore, in order to maximize the amount of light to enter our window would have be
constructed in the shape of a rectangle approximately 5.6 x 2.8 and the semicircular part would
have a radius of approximately 2.8.
5. What is the maximum volume for a rectangular box (square base, no top) made from 12 square feet
of cardboard? (Assume no wasted material)
Let the dimensions of the base be x by x and the height be y.
The function to be maximized is the volume: V = x2y
Since we are constructing the box from 12 ft2 we have the equation:
x2 + 4xy = 12
12  x 2
 y=
4x
 V(x) = x2 *
 V’(x) = 3 -
(**)
1
12  x 2
= 3x - x 3
4
4x
3 2 3
x = (4  x 2 )
4
4
3
(2  x )( 2  x)
4
Which yields critical numbers x = 2 & x = -2 (not applicable)
 V’(x) =
 x=2
Again, let us use the Second Derivative Test
3
12  2 2
V’’(x) = - x  V’’(2) = -3 and we have a Maximum at x = 2 & y =
=1
2
4(2)
The dimensions that yield the maximum volume are 2x2x1 and the Maximum Volume is 4 cubic feet!