i know that if f(x)=f(-x) and if f(-x)=-f(x) then its odd.
i tried to do this check on this cases
but its not working
http://img126.imageshack.us/img126/3942/74448798fv8.gif
i know that if f(x)=f(-x) and if f(-x)=-f(x) then its odd.
i tried to do this check on this cases
but its not working
http://img126.imageshack.us/img126/3942/74448798fv8.gif
for 2A, what is your $\displaystyle -f(x) = -\log \dfrac{1-x}{1+x}$?
2B is an example of a function which is neither odd nor even..
2C.. you haven't done it.. i'll do it for you..
if you replace $\displaystyle x$ with $\displaystyle -x$, then..
$\displaystyle f(-x) = \left\{\begin{array}{cl} -x + 1,& \mbox{for } -x < -1 \\ 0, &\mbox{for }-1\leq -x \leq 1 \\ -x - 1, &\mbox{for } -x > 1 \end{array}\right.$
or
$\displaystyle f(-x) = \left\{\begin{array}{cl} -x + 1,& \mbox{for } x > 1 \\ 0, &\mbox{for }1\geq x \geq -1 \\ -x - 1, &\mbox{for } x < -1 \end{array}\right.$
but this is precisely $\displaystyle -f(x)$.. check it..
$\displaystyle f(x) = \log\left(\frac{1-x}{1+x}\right)$
$\displaystyle f(-x) = \log\left(\frac{1+x}{1-x}\right) = \log\left(\frac{1-x}{1+x}\right)^{-1} = -\log\left(\frac{1-x}{1+x}\right)$
odd function
$\displaystyle f(x) = \sin\left(x + \sqrt{1+x^2}\right)$
$\displaystyle f(-x) = \sin\left(-x + \sqrt{1+(-x)^2}\right) = \sin\left(-x + \sqrt{1+x^2}\right)$
this function and the last one are neither even or odd
well, negation of a piecewise function is negation of the components ONLY of the piecewise function and the negation process does not include the domain of each component..
why don't you negate each component of your function, and you will see that it is precisely the $\displaystyle f(-x)$ i did in the previous post..
well, here it is...
$\displaystyle -f(x) = \left\{\begin{array}{cl} -(x + 1), &\mbox{for } x < -1 \\ 0, &\mbox{for }-1\leq x \leq 1 \\ -(x - 1),& \mbox{for } x > 1 \end{array}\right.$
$\displaystyle f(-x) = \left\{\begin{array}{cl} -x + 1,& \mbox{for } x > 1 \\ 0, &\mbox{for }1\geq x \geq -1 \\ -x - 1, &\mbox{for } x < -1 \end{array}\right.$