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For which $a$ is this function increasing? $$ f(x) = \left( \frac {a-2}{a-4}\right) ^{-x} $$

So first I would rewrite this as: $$ f(x) = \left( \frac {a-4}{a-2}\right) ^{x} $$

I was thinking that in order for the function to be increasing the whole fraction has to be bigger than $1$ or smaller than $-1$

So I devided that into two conditions: $ \frac {a-4}{a-2}> 1 $ and $ \frac {a-4}{a-2} < -1 $

I solved both inequalities and the result should be: for the first inequality: $( -\infty, 2) $
fot the other one: $( 2, 3) $

ANd now for the final result I should combine both so that would be $K =\left\{( -\infty, 2) U ( 2, 3) \right\} $

Is this corrrect? I have no idea how else I should find out .. But my intuition tells me that something is not correct ..

Thanks for help

Lauren Sin
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    Hello - your argument is sound. However, if $\frac{a-4}{a-2} < -1$, then your function is only defined for integer $x$, not for general $x$. For example if $a = \frac{5}{2}$ and $x = \frac{1}{2}$, then $f(x)$ would have to be $\sqrt{-3}$. Stick with the requirement $\frac{a-4}{a-2} > 1$. – Hans Engler Oct 16 '18 at 14:47

2 Answers2

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$\frac {a-4}{a-2}> 1$ is true for all $a<2$

$\frac {a-4}{a-2}<-1$ is true for all $a\in(2,3)$ but not for all real $x$ - consider $x=\frac12$

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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As explained by @Hans, an exponential function is only defined, if the base is positive. And is increasing if the base is greater than $1.$

Thus we solve \begin{equation} \begin{aligned} \frac {a-4}{a-2}&>1\\ \frac{a-4}{a-2}-1&>0\\ \frac{-2}{a-2}&>0\\ \end{aligned} \end{equation} Since the numerator is negative, this is only possible if the denominator $a-2<0.$ The set of solutions is $K =( -\infty, 2).$

user376343
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