Download f(x)=1 x f( 1)=1 ( 1) =1 1=0= f(1) f (x)= 2 3 x f (c)=0 f (0) f 1,1 f(x)=3x +

Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.2 The Mean Value Theorem
2/3
5. f (x)=1 x
2 x
3
/
2/3
. f ( 1)=1 ( 1) =1 1=0= f (1) . f (x)=
1/3
/
, so f (c)=0 has no solution. This
/
does not contradict Rolle's Theorem, since f (0) does not exist, and so f is not differentiable on
( 1,1 ) .
2
11. f (x)=3x +2x+5 , 1,1 . f is continuous on 1,1 and differentiable on ( 1,1 ) since polynomials
f (b) f (a)
f (1) f ( 1) 10 6
/
are continuous and differentiable on R. f (c)=
6c+2=
=
=2 6c=0
b a
1 ( 1)
2
c=0 , which is in ( 1,1 ) .
/
/
15. f (x)= x 1 . f (3) f (0)= 3 1 0 1 =1 . Since f (c)= 1 if c<1 and f (c)=1 if c>1 ,
/
f (c)(3 0)= 3 and so is never equal to 1. This does not contradict the Mean Value Theorem since
/
f (1) does not exist.
3
19. Let f (x)=x 15x+c for x in 2,2 . If f has two real roots a and b in 2,2 , with a<b , then
f (a)= f (b)=0 . Since the polynomial f is continuous on a,b and differentiable on ( a,b) , Rolle's
/
/
2
Theorem implies that there is a number r in ( a,b) such that f (r)=0 . Now f (r)=3r 15 . Since r is
2
in ( a,b) , which is contained in 2,2 , we have r <2 , so r <4. It follows that
2
/
3r 15<3 4 15= 3<0 . This contradicts f (r)=0 , so the given equation can't have two real roots in
2,2 . Hence, it has at most one real root in 2,2 .
25. Suppose that such a function f exists. By the Mean Value Theorem there is a number 0<c<2 with
f (2) f (0) 5
5
/
/
f (c)=
= . But this is impossible since f (x) 2< for all x , so no such function can
2 0
2
2
exist.
29. Let f (x)=sin x and let b<a . Then f (x) is continuous on b,a and differentiable on ( b,a ) . By the
Mean Value Theorem, there is a number c ( b,a ) with
/
sin a sin b= f (a) f (b)= f (c)(a b)=(cos c)(a b) . Thus, sin a sin b cos c b a a b . If
a<b , then sin a sin b = sin b sin a b a = a b . If a=b , both sides of the inequality are 0.
35. Let g(t) and h(t) be the position functions of the two runners and let f (t)=g(t) h(t) . By
hypothesis, f (0)=g(0) h(0)=0 and f (b)=g(b) h(b)=0 , where b is the finishing time. Then by the
f (b) f (0)
/
Mean Value Theorem, there is a time c , with 0<c<b , such that f (c)=
. But f (b)= f (0)=0,
b 0
/
/
/
/
/
/
so f (c)=0 . Since f (c)=g (c) h (c)=0 , we have g (c)=h (c) . So at time c , both runners have the
same speed
1
Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.2 The Mean Value Theorem
/
/
g (c)=h (c).
2