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## anonymous 4 years ago How to do this 1 _______ √1-x^4

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1. campbell_st

is it $1/\sqrt{1-x^4}$

2. anonymous

x^4 has to be less than or equal to 1

3. campbell_st

and what do you have to do..?

4. anonymous

domain maybe?

5. campbell_st

x cannot be 1 or -1

6. anonymous

well the whole problem i will write out

7. anonymous

why can't it be you can have sqrt (0)

8. anonymous

1 is your answer

9. anonymous

well it probably was not √1-4^3

10. anonymous

there was never really a question

11. anonymous

but lemme get the whole problem out first

12. anonymous

oh the 1 is part of the equation..

13. campbell_st

I"ll ray rationalise... $1/\sqrt{1-x^4} \times \sqrt{1 + x^4}/\sqrt{1 +x^4}$ gives $\sqrt{1 - x^4}/(1-x^4)$

14. anonymous

f(x) = x^3 - x^4

15. anonymous

thought he was saying how do you do this one:

16. anonymous

this is the whole problem

17. anonymous

was getting _____ √1-x^4 correct?

18. anonymous

where is that root sign coming from i'm still not sure what you're trying to do

19. anonymous

i just dont know if how i got it was correct entire problem f(x) = x^3 - x^4

20. anonymous

f(x)=x^3(1-x)

21. anonymous

that's just a function you can do ALOT of things with it?

22. anonymous

i am supposed to see if it is even odd or neither

23. anonymous

if it's odd f(x)=-f(-x)

24. campbell_st

well find f(-x) so f(-x) = (-x)^3 - (-x)^4 = -x^3 - x^4 if a function is even f(-x) = f(x) if odd f(-x) = -f(x) in your question the function is neither odd nor even for the above reason

25. anonymous

like he said..

26. anonymous

lol

27. anonymous

tyvm ppls

28. anonymous

:)

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