maths ext question (1 Viewer)

[bettina44]

i just had my maths ext half yearly and there was this question that was bugging me:

show that f(x)= x/(x^2)-1 is an odd function.




when i did it,it show up as neither?

 

[DownInFlames]

i just had my maths ext half yearly and there was this question that was bugging me:

show that f(x)= x/(x^2)-1 is an odd function.




when i did it,it show up as neither?

-f(x) = -x/(x^2 - 1)

f(-x) = -x/(x^2 - 1)


-f(x) = f(-x) therefore odd function

what did you get?

 

[bettina44]

yeah your right,how could i not get that

 

[tommykins]

-f(x) = -x/(x^2 - 1)

f(-x) = -x/(x^2 - 1)


-f(x) = f(-x) therefore odd function

what did you get?

f(-x) = even function by the way.

I believe that's the wrong way to do it (or just a way I've never seen?)

to prove it's an odd function, -f(-x) must yield the same result as f(x)

f(-x) = f(-x) = -x/[(-x)² - 1] = -x/[x² - 1] -> since (-x)² = x²
-f(-x) = - [ -x/(x² - 1) ] = x/(x² - 1) = f(x)

Thus f(x) is an odd function.

 

[Aerath]

Hmmm, what's wrong with the way DowninFlames did it?

 

[The Kaiser]

f(-x) = even function by the way.

I believe that's the wrong way to do it (or just a way I've never seen?)

to prove it's an odd function, -f(-x) must yield the same result as f(x)

f(-x) = f(-x) = -x/[(-x)² - 1] = -x/[x² - 1] -> since (-x)² = x²
-f(-x) = - [ -x/(x² - 1) ] = x/(x² - 1) = f(x)

Thus f(x) is an odd function.

Your way is pretty much the same, just divide both sides of your proof by -1 (since -f(-x) must be the same as f(x) then f(-x) is the same as -f(x))

An easy test is to graph the function (well for me that is) if the right side of the y axis is a mirror image of the left side (that is reflected in both the x and y axis) then the function is odd (eg f(x) = x^3). If the function is even then the left side of the y axis is symetrical in the y axis only. (eg f(x) = x^2)

 

[Devouree]

Your way is pretty much the same, just divide both sides of your proof by -1 (since -f(-x) must be the same as f(x) then f(-x) is the same as -f(x))

An easy test is to graph the function (well for me that is) if the right side of the y axis is a mirror image of the left side (that is reflected in both the x and y axis) then the function is odd (eg f(x) = x^3). If the function is even then the left side of the y axis is symetrical in the y axis only. (eg f(x) = x^2)

with this one it's easier, but generally questions like this in exams are harder in that their almost impossible to sketch accurately. So I'd rather use the algebra method... isn't it -f(x)=f(-x)?

 

[foram]

with this one it's easier, but generally questions like this in exams are harder in that their almost impossible to sketch accurately. So I'd rather use the algebra method... isn't it -f(x)=f(-x)?

Odd function:
f(x)= -f(-x)
or can be written as,
-f(x) = f(-x)

Even function:
f(x) = f(-x)

 

[tommykins]

Ah I see, always preferred to do the f(-x) then -f(-x) as it covers both odd and even whilst you're at it.

 

[DownInFlames]

^^ general maths rules is super duper extremely cool
p.s. you did do general maths downinflames right?

I did 4 unit actually. But I will say that maths in general is super duper extremely cool, and so general maths is super duper extremely cool.

tommykins: my way is the same as your way, only you put both negatives on the same side of the proof.

 

[tommykins]

i've realised, sorry for that mistake

 

[JSchaefer]

even f(x) = f(-x)
odd f(-x) = -f(x)

 

[eliseliselise]

An easy test is to graph the function (well for me that is) if the right side of the y axis is a mirror image of the left side (that is reflected in both the x and y axis) then the function is odd (eg f(x) = x^3). If the function is even then the left side of the y axis is symetrical in the y axis only. (eg f(x) = x^2)

yes, yes!! rotational symmetry-- i normally do the algebraic thing, and then mention rotational symmetry coz i'm a paranoid kid =D