# Question 9c8b1

Mar 16, 2017

x=+-oo

#### Explanation:

Given:$\text{ } \frac{2}{3} {x}^{- \frac{2}{3}} = 0$

Multiply both sides by 3/2

${x}^{- \frac{2}{3}} = 0$

THERE IS A TRAP IN THIS!!!!

It is tempting to say that $x = 0$ This is wrong!

${x}^{- \frac{2}{3}} \text{ is the same as } \frac{1}{{x}^{\frac{2}{3}}}$

Clearly $x \ne 0$ as this is an excluded value in that $\frac{1}{0}$ is 'not allowed'. However as $x \to 0$ then ${x}^{\frac{2}{3}} \to 0 \text{ }$ so $\text{ } \frac{1}{{x}^{\frac{2}{3}}} \to \infty$

So lets consider the other end of the 'number spectrum' instead of x=0 what about $x = \infty$

In this condition we end up with $\frac{1}{{x}^{\frac{2}{3}} \to \infty}$ which is so minute that it may as well be 0, but in this case it exists

So $\frac{2}{3} {x}^{- \frac{2}{3}} = 0 \text{ "=>" } x = \pm \infty$