1. No category

SOLUTIONS SOLUTIONS
1. f(x) = 2x3 + 6x2 – 6x + 7
f '( x)  6 x 2  12 x  6
CALCULUS I
Worksheet #69
SOLUTIONS SOLUTIONS
SOLUTIONS
SOLUTIONS
f "( x)  12 x  12  0  x  1
+ 0 -f "--------|-------2.
Inflection point @ (1,17)
-1
2
y = – x + 4x + 25 on [–2,3]
y '  2 x  4  0  x  2
Critical point @ (2, 29)
Look at endpoints: (-2, 13) and (3, 28)
Maximum y-value is 29
3.
f(x) = x3 – 6x2 on [1,2]
f '( x)  3x 2  12 x  3x( x  4)  0  x  0, 4
Neither critical point is in the given interval, so need to look at endpoints only
(1, -5) and (2, -16)
Absolute minimum y-value is -16.
4.
f ' (x) < 0 and f " (x) > 0
f(x) is decreasing and concave up - A
5
y
1 x
x 3
( x  3)(1)  (1  x)(1)
2
y' 

 2( x  3) 2
2
2
( x  3)
( x  3)
4
y "  4( x  3) 3 
 0  4  0  No inflection point but you must look at x=3
( x  3)3
+ u -y"------------|------------
6.
Concave up (-,3)
3
f(x) = 3x4 + x3
f '( x)  12 x3  3x 2  0  3x 2 (4 x  1)  0  x  0,
-0
+
0 +
f '-----------|--------------|------------1
0
4
1
4
 -1 1 
Rel min @  ,
 ; Terrace pt @ (0,0)
 4 256 
7.
f(x) = 2x3 + 3x2
f '( x)  6 x 2  6 x  6 x( x  1)  0  x  0, 1
+ 0 -- 0 +
f '    |      |     
-1
Increasing (-, -1) & (0, )
0
2 x  y  160  y  160  2 x
8.
Building – no fence here
A  xy  x(160  2 x)  160 x  2 x 2
A '  160  4 x  0  x  40; y  80
x
x
y
9.
f ' (x) = 0 and f " (x) > 0
10.
f ' (x) > 0 and f " (x) > 0
Critical point and concave up  E
Increasing and concave up  B
--
11.
f '(9)  0
f '      |       This means f(x) is decreasing at x = 9  C
9
Not enough info to determine whether I and II are true all the time.
12.
y
x 2  3x  2 ( x  2)( x  1)
x2

y
2
x  4 x  3 ( x  3)( x  1)
x 3
 1
Hole: 1, 
 2
Vertical Asymptote: x = 3
Horizontal Asymptote: y = 1
-2
 2
y-intercept (where x=0) y 
  0, 
-3
 3
x-intercept (where y = 0)  0 

x-2
 0  x  2  x  2  (2, 0)
x -3
y




x












13.
x2
F ( x)   3sint 2 dt  F '( x)  3sin( x 2 ) 2 (2 x)  6 x sin x 4
4
.
14.
x = 3y – y2 and y = x; revolved around the y-axis
Washer
R  3 y  y 2 and r = x
56
V    [(3 y  y )  y ]dy 
15
0
2
2 2
15.
2
y




x









Base: x2 + y2 = 16. Cross sections: perpendicular to the y=axis are squares.
x 2  16  y 2
x   16  y 2
x   16  y 2
s  16  y 2   16  y 2  2 16  y 2
 2
V
4
 dy  1024
3
2
4
16  y
2
s
x  16  y 2