How to do this 1
_______
√1-x^4

Question

How to do this 1
  _______
√1-x^4

campbell_st · Answer

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

anonymous · Answer

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

campbell_st · Answer

and what do you have to do..?

anonymous · Answer

domain maybe?

campbell_st · Answer

x cannot be 1 or -1

anonymous · Answer

well the whole problem  i will write out

anonymous · Answer

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

anonymous · Answer

1 is your answer

anonymous · Answer

well it probably was not √1-4^3

anonymous · Answer

there was never really a question

anonymous · Answer

but lemme get the whole problem out first

anonymous · Answer

oh the 1 is part of the equation..

campbell_st · Answer

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

anonymous · Answer

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

anonymous · Answer

thought he was saying how do you do this one:

anonymous · Answer

this is the whole problem

anonymous · Answer

was getting 
   _____
√1-x^4

correct?

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

campbell_st · Answer

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

anonymous · Answer

like he said..

anonymous · Answer

lol

anonymous · Answer

tyvm ppls

anonymous · Answer

:)