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Let g be a continuous function on the closed interval [0,1].
Let g(0)=1 g(1)=0
Which of the following is NOT necessarily true?

a) There exists a number h in [0,1] such that g(h) > or = g(x) for all x in [0,1].

b) For all a and b in [0,1], if a=b, then g(a) = g(b)

c) There exists a number h in [0,1] such that g(h)= 0.5

d) There exists a number h in [0,1] such that g(h) = 3/2

e) For all h in the open interval (0,1), the limit as x goes to h of
g(x) = g(h)

and why.


thanks so much

2006-09-24 17:46:08 · 5 answers · asked by leksa27 2 in Science & Mathematics Mathematics

5 answers

a) true, because a continuous image of a compact interval is compact, hence bounded, so there is an upper bound.

b) true, because it is a function.

c) true, because a continuous image of a convex interval is convex; the image contains 0 and 1 and therefore every number in between.

d) not true; 3/2 does not lie in the interval [0,1]. A counterexample is the function g(x) = 1 - x.

e) true, because in the interior of the domain of a continuous function all limits are equal to the value of the function in that point

2006-09-24 17:53:00 · answer #1 · answered by dutch_prof 4 · 0 0

a is true, any finite function that is continuous and is bounded has a max and min.

b is true for any function (example let a=.25 then if a=b b=.25 and f(b)=f(.25)=f(a)

c is true by the mean value theorem.

d is not always true

e I'm pretty sure this one is true too, i think this is part of the definition of a continuous function

2006-09-24 17:57:35 · answer #2 · answered by sparrowhawk 4 · 0 0

all are true except d

for example g(x) = 1 -x^2 satisfies the given condition and for no h is g(h) = 3/2

2006-09-24 17:53:51 · answer #3 · answered by Mein Hoon Na 7 · 0 0

I'm curious too

2016-08-08 15:48:18 · answer #4 · answered by Anonymous · 0 0

It depends..

2016-08-23 07:31:39 · answer #5 · answered by Anonymous · 0 0

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