1) Is it possible to have a function f defined on [ 2 , 4 ] and meets the given conditions?
f is continuous on [ 2 , 4 ), minimum value f(4)=2, and no maximum value.
2) Is it possible to have a function f defined on [ 4 , 5 ] and meets the given conditions?
f is continuous on [ 4 , 5 ], takes on no rational values.
3) Is it possible to have a function f defined on [ 2 , 5 ] and meets the given conditions?
f is continuous on [ 2 ,5 ] and the range of f is an unbounded interval.

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- clara1223

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- anonymous

i am thinking of a counter example for the first one, but i can't seem to come up with one
i bet you can come up with and example for #2, think of a constant function

- anonymous

#3 contradicts whatever theorem it is that says a continuous function on a closed interval has a max and a min

- anonymous

i bet @Zarkon can come up with an example for the first one

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## More answers

- clara1223

So far we have that for 2 it is possible, but for 3 it isn't possible?

- anonymous

ooh i got one

- Zarkon

|dw:1441940193237:dw|

- anonymous

who said art was dead?

- clara1223

that makes a lot of sense!

- clara1223

so 1 is possible, 2 is possible, but 3 isn't possible. correct?

- anonymous

i was thinking of \[\frac{1}{4-x},2\] or something similar |dw:1441940311469:dw||dw:1441940327769:dw|

- anonymous

yeah did you come up with an example for #2?

- clara1223

|dw:1441940405617:dw|

- anonymous

what is \(y\)?

- clara1223

I didn't think it mattered, as long as x=4 and x=5 were defined

- anonymous

it matters a great deal
for example, if your function is
\[f(x)=3\]then it is rational on that inteval

- clara1223

so what y value would satisfy this?

- anonymous

btw defined on \([4,5]\) means defined for all \(x\) in the interval \(4\leq x\leq 5\)not just at the endpoints

- anonymous

why doesn't \[f(x)=3\]work?

- clara1223

i'm confused

- anonymous

\[f(x)=3\]is an example of a function that is continuous and defined on any interval you like, including \([4,5]\) but it s not always irrational on that interval because... because it is always rational on the interval on account of 3 is a rational number

- anonymous

so that is NOT an example, but you can modify it so that it is an example of a function that is never rational
just don't pick 3

- clara1223

oh, so would y=sqrt(2) work? that is an irrational number

- anonymous

yup

- clara1223

cool! thanks

- anonymous

yw

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