# Thread: First derivative test, turning points

1. ## First derivative test, turning points

I don't understand how the LHS equals the RHS for this:

1 - x^(-2) = (x^2 - 1)/x^2

Then this statement follows which i also don't understand.

"Turning points when dy/dx = 0 are when x^2 - 1 = 0, that is when x = -1, 1.

Note that dy/dx is undefined at x = 0."

Any help would be much appreciated.

2. Originally Posted by anon_404
I don't understand how the LHS equals the RHS for this:

1 - x^(-2) = (x^2 - 1)/x^2

Then this statement follows which i also don't understand.

"Turning points when dy/dx = 0 are when x^2 - 1 = 0, that is when x = -1, 1.

Note that dy/dx is undefined at x = 0."

Any help would be much appreciated.
$x^{-2} = \frac{1}{x^2}$

To get the LHS to equal the RHS multiply the 1 by x^2 and combine with the same denominator:

$1 \times \frac{x^2}{x^2} - \frac{1}{x^2} = \frac{x^2-1}{x^2}$

To find the turning points set your equation above to 0 and solve. I assume the above equation is the derivative otherwise that statement is false.

It is undefined at x=0 because you can't divide by 0 and the x^2 on the bottom would be 0

3. Originally Posted by anon_404
I don't understand how the LHS equals the RHS for this:

1 - x^(-2) = (x^2 - 1)/x^2

Then this statement follows which i also don't understand.

"Turning points when dy/dx = 0 are when x^2 - 1 = 0, that is when x = -1, 1.

Note that dy/dx is undefined at x = 0."

Any help would be much appreciated.

Hey in terms of the first part:

x^(-2) is the same as 1/x^2. (RULES OF INDICES)

So on the LHS you have:

1- (1/x^2).

So you have:

(x^2/x^2) - (1/x^2) = (x^2 - 1)/ x^2 (FRACTION RULES)

Sorry this answer is a bit crap. I'm not that good at explaining things.

4. Originally Posted by e^(i*pi)

To get the LHS to equal the RHS multiply the 1 by x^2 and combine with the same denominator:

$1 \times \frac{x^2}{x^2} - \frac{1}{x^2} = \frac{x^2-1}{x^2}$

Would you be able to explain the fraction part more. Sorry i have little algebra knowledge

5. Originally Posted by anon_404
I don't understand how the LHS equals the RHS for this:

1 - x^(-2) = (x^2 - 1)/x^2
The RHS is $\frac{x^2- 1}{x^2}= \frac{x^2}{x^2}- \frac{1}{x^2}$
$= 1- \frac{1}{x^2}= 1- x^{-2}$

Then this statement follows which i also don't understand.

"Turning points when dy/dx = 0 are when x^2 - 1 = 0, that is when x = -1, 1.

Note that dy/dx is undefined at x = 0."

Any help would be much appreciated.
"Turning points" of a graph must occur where the derivative is 0. A fraction is 0 only where the numerator of is 0. That is, $f'(x)= \frac{x^2- 1}{x^2}= 0$ when $x^2- 1= 0$ or $x^2= 1$ so x= 1 or x= -1.

Of course, division by 0 is undefined so $f'(x)= \frac{x^2-1}{x^2}$ is undefined when the denominator, $x^2$, is 0: $x^2= 0$ so x= 0.