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(A) There exists a $y$ in the interval $(0,1)$ such that $f(y)=f(y+1)$.

(B) For every $y$ in the interval $(0,1), f(y) = f(2−y)$.

(C) The maximum value of the function in the interval $(0,2)$ is $1$.

(D) There exists a $y$ in the interval $(0,1)$ such that $f(y) = − f(2−y)$.

How to approach such questions ?

JKnecht
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radhika
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    Look for counterexamples, and use the theorems in your textbook. – Christopher Carl Heckman Feb 01 '16 at 07:35
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    You approach such questions by trying stuff. Trying to prove or disprove every point. And yes, you have to do it and fail at it. Several times. Asking people to do it is not how you learn math, – 5xum Feb 01 '16 at 07:37
  • If I take y=0.8 ,Now how to check what is f(1.8) ? then only I can actually calculate what is f(y+1) ? – radhika Feb 01 '16 at 07:37
  • The questions asked in the title and in the post are different. Based on the title, this should be closed for lack of context. The question asked in the post is fully answered by @CarlHeckman's comment. – Did Feb 01 '16 at 07:38
  • Does $f(0.8)$ HAVE to equal $f(1.8)$? Can you create a continuous function such that $f(0.8)=1$ and $f(1.8)=0$, for instance? – Christopher Carl Heckman Feb 01 '16 at 07:38

1 Answers1

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For all cases (B-D) the counter-example is very easy. And I skip them.

I just prove case A:

define

$$h(t)=f(t+1)-f(t)$$

$$h(0)=2 ~~~~\text{and}~~~~ h(1)=-2$$

$f$ is continuous so as $h$, so there is a $0<y<1$ where $h(y)=0$ or $f(y)=f(y+1)$

Arashium
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