Find Least Value Inequality for $$-1<x<0$$

[Amad27]

Which of the following have the least value if

$$-1 < x < 0$$

$$(A) -x$$
$$(B) 1/x$$
$$(C) -1/x$$
$$(D) 1/x^2 $$
$$(E) 1/x^3$$

Mmmmmmmm...

I'm not sure what to do, but I'll definitely try. We can break it up into two inequalities.

$$ x > -1$$
$$0 > x$$

$$\implies -x < 1, 0 < -x$$
$$\implies -1 < 1/x, 0 < 1/x$$
$$\implies 1 > 1/x, 0 > -1/x$$
$$\implies 1/x^2 < 1$$
$$ \implies 1/x^3 < -1$$

So $(E)$ should be correct.

BottomLine: Is this the correct way to go about it?

Thanks!

 

[earboth]

Which of the following have the least value if

$$-1 < x < 0$$

$$(A) -x$$
$$(B) 1/x$$
$$(C) -1/x$$
$$(D) 1/x^2 $$
$$(E) 1/x^3$$

Mmmmmmmm...

I'm not sure what to do, but I'll definitely try. We can break it up into two inequalities.

$$ x > -1$$
$$0 > x$$

$$\implies -x < 1, 0 < -x$$
$$\implies -1 < 1/x, 0 < 1/x$$
$$\implies 1 > 1/x, 0 > -1/x$$
$$\implies 1/x^2 < 1$$
$$ \implies 1/x^3 < -1$$

So $(E)$ should be correct.

BottomLine: Is this the correct way to go about it?

Thanks!

Hej + good morning,

Is this the correct way to go about it?

as usual in math there isn't the one and only way to do a question: If you get the correct result your way is probably correct too.

Here is how I would have done this question:
Let $$k \in \mathbb{R}\ \wedge \ k > 1~\implies~x = -\frac1k$$

If you look for the least value of a term all terms which produce a positive result are not valid. So that leaves the B and E. Now replace x by $$-\frac1k$$ and you'll see immediately that $$\underbrace{-k^3}_{\text{case E}} < \underbrace{-k}_{\text{case B}}$$