Exercises 1 8.3 27. y = f ( x) = f ( x) = = -2
(
2
1 . x + 2x + 5 x 2 + 2 x + 5 0 - 1 ( 2 x + 2)
2
. 2 + 2x + 5 f ( x) = 0 for x = -1. f ( x) > 0 for x < -1, f ( x) < 0 for x > -1. 1 Maximum value of f(x) is f(-1) = . 4
(x
x +1
(x
)
2
+ 2x + 5
)
2
.
)
(x f ( x) = -2 (x (x (x
2 2
2
+ 2 x + 5 1 - ( x + 1) 2( x 2 + 2 x + 5)(2 x + 2)
)
2
= -2 = -2 = -2 = 2
+ 2 x + 5 - 4( x + 1) 2
(x
) )
(x
2
+ 2x + 5
)
4
2
2
+ 2x + 5
2
)
3
2
+ 2 x + 5 - 4( x 2 + 2 x + 1)
(x
+ 2x + 5
)
3
- 3x 2 - 6 x + 1
2
+ 2x + 5
)
3
+ 2x + 5 f ( x) = 0 for :
(x
3x 2 + 6 x - 1
)
3
3 x 2 + 6 x - 1 = 0. x= x x x x - 6 6 2 - 4 3 (-1) . 23 - 6 48 = . 6 -64 3 = . 6 2 = -1 . 3 = .1547 or x = -2.1547. 2 = -2.1547. 3
f ( x) > 0 for x < -1 -
2
2 2 < x < -1 + -2.1547 < x < .1547. 3 3 2 f ( x) > 0 for x > -1 + = .1547. 3 f ( x) < 0 for - 1 Here R is the region of the (x, y) - plane bounded by the graph of y = f(x) = 1 , x + 2x + 5
2
the x - axis, the y - axis, and the line x = 1. Let S be the solid of revolution generated by revolving R about the x - axis. We are to find the volume of S. On the interval 0 x 1 : y = f(x) > 0 f (x) < 0, so f(x) is decreasing. 1 1 f(0) = and f(1) = . 5 8 For 0 < x < .1547, f (x) < 0, so the graph is concave downward. For .1547, (x) f > 0, so the graph is concave upward. The point (.1547, f(.1547)) is an inflection point. Solve the equation 1 x + 2x + 5 for x in terms of y. 1 = x 2 + 2 x + 5. y y=
2
1 x 2 + 2 x + 5 - = 0. y 1 - 2 22 - 4 5 - y x= . 2 1 1 - 2 4 - 16 y x= . 2 1 x = -1 - 4. y
3
Take x = -1 + 1 - 4. y
2
The volume of S is given by [ f(x)] dx
1 0
= =
1
0 1
(x
1
2
0
Let x + 1 = 2tan ( ). Then
( ( x + 1)
+ 2x + 5 1
2
)
2
dx dx.
+ 22
)
2
( x + 1) 2 + 2 2 = 4 tan 2 ( ) + 4 = 4 ( tan 2 ( ) + 1) = 4 sec 2 ( ).
Thus Also dx = 2sec 2 ( ) d . Hence
(( x + 1)
2
+ 22
)
2
= 16 sec 4 ( ).
(( x + 1)
=
1
2
+ 22
)
2
dx
1 2sec 2 ( ) d 4 16 sec ( ) 1 = cos 2 ( ) d 8 1 1 + cos( 2 ) = d 8 2 1 1 = + sin ( 2 ) 16 2 1 = ( + sin ( ) cos( ) ). 16 x 2 + 2x + 5 2
x+...