If f is continuous on [0,2] with f(0) = f(2), show that there is a c element of [0,2] such that f(c) = f(c+1)?

Answer 1

We consider the following function #g(x)=f(x+1)-f(x)#

Hence we have that

#g(1)=f(2)-f(1)=f(0)-f(1)# and
#g(0)=f(1)-f(0)#

If we take the product we have that

#g(0)*g(1)=-(f(0)-f(1))^2<0#
From Intermediate value theorem that means there is a c for which #g(c)=0=>f(c)=f(c+1)# qed
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Answer 2

To show that there exists a ( c ) in the interval ([0,2]) such that ( f(c) = f(c+1) ), you can consider the function ( g(x) = f(x) - f(x+1) ). Since ( f(x) ) is continuous on ([0,2]), so is ( g(x) ).

Evaluate ( g(0) ) and ( g(1) ): [ g(0) = f(0) - f(1) ] [ g(1) = f(1) - f(2) ]

Given ( f(0) = f(2) ), we have ( g(0) = -g(1) ).

By the Intermediate Value Theorem, since ( g(x) ) is continuous on ([0,1]) and ( g(0) = -g(1) ), there exists a ( c ) in ([0,1]) such that ( g(c) = 0 ).

Thus, ( f(c) = f(c+1) ), where ( c ) is in the interval ([0,1]), which is a subset of ([0,2]).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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