Thomas Calculus 13th [Solutions]

234 Chapter 4 Applications of Derivatives
17. (a)
n n 1
(b) Let r 1 and r 2 be zeros of the polynomial P( x) x an
1x a1 x a 0,
then P( r1 ) P( r2
) 0.
Since polynomials are everywhere continuous and differentiable, by Rolles Theorem P ( r ) 0 for some r
n 1
n 2
between r 1 and r 2 , where P ( x)
nx ( n 1) an
1x a 1 .
18. With f both differentiable and continuous on [ a, b ] and f ( r1 ) f ( r2 ) f ( r 3) 0 where r 1 , r 2 and r 3 are in [ a , b ],
then by Rolles Theorem there exists a c 1 between r 1 and r 2 such that f ( c 1) 0 and a c 2 between r 2 and r 3 such
that f ( c 2) 0. Since f is both differentiable and continuous on [ a, b ], Rolles Theorem again applies and we
( n)
have a c 3 between c 1 and c 2 such that f ( c 3) 0. To generalize, if f has n 1 zeros in [ a, b ] and f is continuous
( n)
on [ a, b ], then f has at least one zero between a and b.
19. Since f exists throughout [ a, b ] the derivative function f is continuous there. If f has more than one zero
in [ a, b ], say f ( r1 ) f ( r 2) 0 for r 1 r 2 , then by Rolles Theorem there is a c between r 1 and r 2 such that
f ( c ) 0, contrary to f 0 throughout [ a, b ]. Therefore f has at most one zero in [ a, b ]. The same argument
holds if f 0 throughout [ a, b].
20. If f ( x ) is a cubic polynomial with four or more zeros, then by Rolles Theorem f ( x ) has three or more zeros,
f ( x ) has 2 or more zeros and f ( x ) has at least one zero. This is a contradiction since f ( x ) is a non-zero
constant when f ( x ) is a cubic polynomial.
21. With f ( 2) 11 0 and f ( 1) 1 0 we conclude from the Intermediate Value Theorem that
4
3
3
f ( x) x 3x 1 has at least one zero between 2 and 1. Then 2 x 1 8 x 1 32 4x
4
3
29 4x 3 1 f ( x ) 0 for 2 x 1 f ( x ) is decreasing on [ 2, 1] f ( x ) 0 has exactly one
solution in the interval ( 2, 1).
22. f 3 2
( x ) x 4 7 f ( x ) 3 x 8 0 on ( , 0) f ( x ) is increasing on ( , 0). Also, f ( x ) 0 if x 2
2 3
x
and f ( x ) 0 if 2 x 0 f ( x ) has exactly one zero in ( , 0).
x
23. g( t) t t 1 4 g ( t ) 1 1 0 g ( t ) is increasing for t in (0, ); g (3) 3 2 0 and
2 t 2 t 1
g(15) 15 0 g( t ) has exactly one zero in (0, ).
24. g( t) 1 1 t 3.1 g ( t)
1 1 0 g( t ) is increasing for t in ( 1,1); g ( 0.99) 2.5 and
1 t
2
(1 t)
2 1 t
g(0.99) 98.3 g( t ) has exactly one zero in ( 1, 1).
25.
26.
2
r ( ) sin 8 r ( ) 1 2 sin cos 1 1 sin 2 0 on ( , ) r ( ) is increasing on
3
3 3 3 3 3
2
( , ); r (0) 8 and r (8) sin 8 0 r ( ) has exactly one zero in ( , ).
3
2
r ( ) 2 cos 2 r ( ) 2 2sin cos 2 sin 2 0 on ( , ) r ( ) is increasing on
( , ); r ( 2 ) 4 cos( 2 ) 2 4 1 2 0 and r(2 ) 4 1 2 0 r ( ) has exactly one
zero in ( , ).
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