# Math Help - Locating stationary points on a curve

1. ## Locating stationary points on a curve

The question:

Locate and identify the stationary points for $y = \frac{x}{1+x^2}$

My attempt:

Using quotient rule...
$\frac{1 - x^2}{(1 + x^2)^2}$

I find when this equals zero, i.e. $1 - x^2 = 0$
Therefore, $x = \pm1$

I substitute this into the original eqn to get 1/2 and -1/2. So the points are (1, 1/2) and (-1, -1/2). So far so good.

Now I want to know the nature of these points. So I go to find the second derivative...

$\frac{-2x(1+x^2)^2 - (1 - x^2)4x(1 + x^2)}{(1 + x^2)^4}$

Looks a bit messy!

I substitute 1 into the equation and get a negative number, so I gather that (1, 1/2) is a maximum point. However, when I substitute -1, I get 0. My text states that (-1, -1/2) is a local minimum, so it can't be 0, can it?

Any assistance would be greatly appreciated!

2. I'm not sure why you substituted $\pm 1$ in the second derivative.

But substituting $\pm 1$ in $\frac{x}{1+x^2}$ gives points (1,1/2) and (-1,-1/2)

We know these points are either local or global maxima/minima.
We only have to check whether there exist $x\in \mathbb{R}$ such that $y(x)>1/2$ or $y(x)< -1/2$

(if you check the graph of y(x) on wolframalpha we actually see that (1,1/2) and (-1,-1/2) are global maxima/minima. Thus I don't understand why your
text states (-1,-1/2) is a local minimum. )

3. Originally Posted by Dinkydoe
I'm not sure why you substituted $\pm 1$ in the second derivative.
I was trying to do this:
Second derivative test - Wikipedia, the free encyclopedia

4. Originally Posted by Glitch
I haven't checked any of your calculations but if you're using the second derivative test you should have made it your business to know that the test fails if f'' = 0. In such cases the sign test must be used. Regardless, for your problem it's much more sensible to just use the sign test in the first place.

5. Originally Posted by mr fantastic
I haven't checked any of your calculations but if you're using the second derivative test you should have made it your business to know that the test fails if f'' = 0. In such cases the sign test must be used. Regardless, for your problem it's much more sensible to just use the sign test in the first place.
Ok, I just did it, and it is indeed a minimum point. Thanks, I should have known to check that way. The second derivative is very messy, after all.

6. Originally Posted by Glitch
The question:

Locate and identify the stationary points for $y = \frac{x}{1+x^2}$

My attempt:

Using quotient rule...
$\frac{1 - x^2}{(1 + x^2)^2}$

I find when this equals zero, i.e. $1 - x^2 = 0$
Therefore, $x = \pm1$

I substitute this into the original eqn to get 1/2 and -1/2. So the points are (1, 1/2) and (-1, -1/2). So far so good.

Now I want to know the nature of these points. So I go to find the second derivative...

$\frac{-2x(1+x^2)^2 - (1 - x^2)4x(1 + x^2)}{(1 + x^2)^4}$

Looks a bit messy!

I substitute 1 into the equation and get a negative number, so I gather that (1, 1/2) is a maximum point. However, when I substitute -1, I get 0.
No, you don't! When x= -1, this becomes
$\frac{-2(-1)(1+(-1)^2)^2 - (1 - (-1)^2)4(-1)(1 + (-1)^2)}{(1 + (-1)^2)^4}$
$\frac{(2)(4)}{16}= \frac{1}{2}> 0$.

I wonder it you neglected squaring -1 before adding it to 1?

My text states that (-1, -1/2) is a local minimum, so it can't be 0, can it?

Any assistance would be greatly appreciated!

7. Originally Posted by HallsofIvy
No, you don't! When x= -1, this becomes
$\frac{-2(-1)(1+(-1)^2)^2 - (1 - (-1)^2)4(-1)(1 + (-1)^2)}{(1 + (-1)^2)^4}$
$\frac{(2)(4)}{16}= \frac{1}{2}> 0$.

I wonder it you neglected squaring -1 before adding it to 1?
Hmm, I made sure I put brackets on my calculator to ensure that it would square properly. Perhaps I made a typo. Thanks.

8. You can always use this mnemonic device for remembering the Second Derivative Test:

If the second derivative is negative (-), you're concave down (first face). If the second derivative is positive (+), you're concave up (second face). If the second derivative is zero (0), you could be concave down, concave up, or have an inflection point (third, confused face).

Cheers!