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I want to show these properties with two functions:

  • even: $f(x)=f(-x)$
  • odd: $-f(x)=f(-x)$

Prove that the function $g: D \to \mathbb{R}$, $g(t)= \frac {5} {t^4 - t^2 + 1}$ is even and $h: \mathbb{R} \{0 \} \to \mathbb{R}$, $h(a)= \frac {1+a^2} {a} is odd$

I tried to prove it by:

$$g(2)= -\frac {5} {2^4 - 2^2 + 1} = -\frac {5} {13}$$

Therefore the first function is not even

The second function is:

$$-h(1)= -\frac {1+1^2} {1} = -2$$

Which shows that it is not odd.

Are my assumptions correct?

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    $\begingroup$ You need to prove that $g(t) = g(-t)$ for every $t$ in $D$. Start by substituting $t$ by $-t$ in the given formula for $g(t)$. $\endgroup$ – Anthony Carapetis Aug 7 '13 at 7:40
  • $\begingroup$ @AnthonyCarapetis By replacing I get $g(-t)= \frac {5} {-t^4 + t^2 + 1}$ However, how does this show that $g(t)=g(-t)$ $\endgroup$ – user2051347 Aug 7 '13 at 7:50
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    $\begingroup$ Be careful! $-(t^2) \ne (-t)^2$. $\endgroup$ – Anthony Carapetis Aug 7 '13 at 7:54
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Using your example:

$$g(2)=\frac5{2^4-2^2+1}=\frac5{(-2)^4-(-2)^2+1}=g(-2)$$

and, in fact:

$$g(x)=\frac5{x^4-x^2+1}=\frac5{(-x)^4-(-x)^2+1}=g(-x)\iff \;g\;\;\text{is an even function}$$

But

$$h(a)=\frac{1+a^2}a=-\frac{1+a^2}{-a}=-h(-a)\;,\;\;\text{so}\ldots$$

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  • $\begingroup$ Thx for your answer! In fact you just multiplied $\frac5{x^4-x^2+1}$ by $-1$ and proved it like that $\ddot\smile$ $\endgroup$ – user2051347 Aug 7 '13 at 7:53
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    $\begingroup$ You're welcome...but no: I didn't multiply that by $\,-1\,$. What I did is to show that inputting $\,x\,$ and inputting $\;-x\;$ yields the same result. $\endgroup$ – DonAntonio Aug 7 '13 at 7:54
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Your assumptions are not correct. You can't prove $f(x)$ is even by disproving that $f(x)$ is odd. There are many functions that are neither (e.g. $e^x$).

To prove that g(t) is odd you need to start with the left hand side of the definition of odd (i.e. $-g(x)$) and manipulate it in a way so you arrive at the right hand side of that equation (i.e. $f(-x)$).

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