## anonymous one year ago 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.

1. 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

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

ooh i got one

6. Zarkon

|dw:1441940193237:dw|

7. anonymous

8. anonymous

that makes a lot of sense!

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

|dw:1441940405617:dw|

13. anonymous

what is $$y$$?

14. anonymous

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

15. anonymous

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

16. anonymous

so what y value would satisfy this?

17. anonymous

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

18. anonymous

why doesn't $f(x)=3$work?

19. anonymous

i'm confused

20. 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

21. 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

22. anonymous

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

23. anonymous

yup

24. anonymous

cool! thanks

25. anonymous

yw

Find more explanations on OpenStudy