3

Given, for example:

$$ \Big(\frac{2x}{x-2}\Big)^{3x^2-x} \leq \Big(\frac{2x}{x-2}\Big)^{x^2+3x+6} $$

After checking when $x-2 \ne 0$ , the teacher taught us to check 3 cases:
1. $ \Big(\frac{2x}{x-2}\Big)> 1 $ , and then the inequality sign remains the same
2. $ 0 < \Big(\frac{2x}{x-2}\Big) < 1 $ , and then the inequality sign flips
3. $ \Big(\frac{2x}{x-2}\Big) = 1 $ , and then add the solution (if there is one) at the end

My question is, why don't we need to check for more cases, like: $ \Big(\frac{2x}{x-2}\Big) = 0 $ ?
Or: $ -1 < \Big(\frac{2x}{x-2}\Big) < 0 $ ?
Or: $ \Big(\frac{2x}{x-2}\Big) < 1 $ ?

Aren't those cases also relevant?

Ken White
  • 31
  • 1
    It's because for a non-integer exponent, the expression under the exponent has to be positive (it's defined with \ln). – Bernard Aug 26 '15 at 22:25
  • @Bernard - without trying to complicate things, what if the exponent is a rational number $\frac{p}{q}$ in lowest terms with $q$ odd, e.g. when $x=-\frac{1}{5}$? – Marconius Aug 26 '15 at 22:34
  • There are issues with $a^b$ when $a$ is negative. For example, what is $(-1)^\pi$? But there is a good case to be made that the inequality makes sense when, for example, $x=1$, that is, $\frac{2x}{x-2}=-2$. – André Nicolas Aug 26 '15 at 22:43
  • @Marconius: While we can speak of a root of odd order for a negative number, we do not use usually fractionary exponents for these roots, as using the usual formulae for exponents would lead to contradictions. For instance $(-8)^{\tfrac13}=-2=(-8)^{\tfrac2{6}}=\bigl((-8)^2\bigr)^{\tfrac16}=2$. Of course, this is due to the fact that a rational number doesn't have a single representation as a fraction. – Bernard Aug 26 '15 at 22:54

0 Answers0