# Thread: Find stationery point where x to -ve power. How?

1. ## Find stationery point where x to -ve power. How?

Hi

To find stationery points I always differentiate the function which in my previouas work has always resulted in a quadratic equation so set to zero and factorise or find roots.

But what to do if gradient is eg x^-2 ???

the function is 4x^2 + 1/x

dy/dx = 8x - x^-2

8x - x^-2 = 0 ??? How do I solve this?

Angus

2. Originally Posted by angypangy
To find stationery points I always differentiate the function which in my previouas work has always resulted in a quadratic equation so set to zero and factorise or find roots.
But what to do if gradient is eg x^-2 ???
the function is 4x^2 + 1/x
dy/dx = 8x - x^-2
8x - x^-2 = 0 ??? How do I solve this?
$\displaystyle 8x-x^{-2}=\frac{8x^3-1}{x^2}$.

3. Sorry I am probably being a bit thick but any more hints? I am not understanding your point?

4. Or: $\displaystyle 8x- x^{-2}= 0$ is the same as $\displaystyle 8x= x^{-2}$ so, multiplying both sides by $\displaystyle x^2$, $\displaystyle 8x^3= 1$. That is the same as $\displaystyle x^3= \frac{1}{8}$

5. Are you saying that you cannot solve $\displaystyle \frac{8x^3-1}{x^2}=0~?$.

6. Sorry I was confused with something I read about the quotient rule which I have not yet covered and I thought I needed to use somehow. Yes understand now.