Thomas Calculus 13th [Solutions]

Chapter 4 Practice Exercises 335
x
(c) y (1 x)
e
x x x
y e (1 x)
e xe
x x x
y e e ( x 1) e ; solving
x
y 0 xe 0 x = 0; y 0 for
x > 0 and y 0 for x < 0 a maximum at x =
0
0 of (1 0) e 1; there is a point of inflection
at x = 1 and the curve is concave up for x > 1
and concave down for x < 1.
134. y = x ln x y ln x x 1 ln x 1; solving
x
y 0 ln x + 1 = 0 ln x = 1
y 0 for
1
x e and y 0 for
1 1
a minimum of ln 1
1
e e at
e x e . This
minimum is an absolute minimum since y 1 is
x
positive for all x > 0.
x
x
e
1
e
1 ;
135. In the interval < x < 2 the function
sin x
sin x < 0 (sin x ) is not defined for all values
in that interval or its translation by 2 .
2 2 2
136. v x ln 1 x (ln1 ln x) x ln x dv
2
2x ln x x 1 x(2ln x 1); solving
x
dx
x
dv
1 1/2
0 2ln x 1 0 ln x x e ; dv
1/2
0 for x e and dv
1/2
0 for x e a relative
dx
2
dx
dx
1/2
maximum at x e ; r
1/2
x and r = 1 h e e 1.65 cm
h
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