Is 1 / ( x )^2 strictly greater than  1/(x) for all x?

Oct 22, 2015

No

Explanation:

$\frac{1}{x} ^ 2 \le \frac{1}{x} \mathmr{if} x \ge 1$
$\textcolor{w h i t e}{\text{XXX}}$for example $\frac{1}{{2}^{2}} = \frac{1}{4} < \frac{1}{2}$

Oct 22, 2015

See explanation.

Explanation:

If $x < 0$:

When $x$ is negative, $\frac{1}{x}$ will be negative, but $\frac{1}{x} ^ 2$ will always be positive (because if $x \in \mathbb{R}$, then ${x}^{2} \ge 0$). Therfore, $\frac{1}{x} ^ 2$ will be greater than $\frac{1}{x}$.

If $x = 0$:

Both expressions will be undefined. Therefore, you can't compare them.

If $0 \text{<} x < 1$:

When $x$ is between 0 and 1, $\frac{1}{x} ^ 2$ will be greater than $\frac{1}{x}$.

If $x = 1$:

Both expressions will be equal (they will both be 1).

If $x > 1$:

When $x$ is greater than 1, $\frac{1}{x}$ will be greater than $\frac{1}{x} ^ 2$. This is because in $\frac{1}{x}$, you are dividing 1 by $x$ only once, while in $\frac{1}{x} ^ 2$, you are dividing 1 by $x$ twice.