Why are 1st, 2nd derivative tests giving different answers? y'=(-x^2-6x-5)/(x2+x-2)^2

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hndalama

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Find the nature of the turning points? why are these two alternative methods giving me different answers.
Dy/dx= (-x2 -6x -5)/(x2+x-2)2
(-x2 -6x -5)/(x2+x-2)2 =0
-x2 -6x -5 =0
X2+6x+5=0
(x+5)(x+1) = 0
So turning points at x = -1 or -5
First derivative test
Track gradient around critical points
The denominator is squared so the denominator will always be positive hence I only consider the numerator i.e. (-x2-6x-5).
X
-2
-1
0
-5
-6
Gradient sign
+
0
-
0
-

at x=-5, the gradient stays –ve indicating that it is an inflection point.
However when I solve by finding the second derivative of dy/dx I get the following:
Dy/dx= (-x2 -6x -5)/(x2+x-2)2
D2y/dx2=[ (-2x-6) (x2+x-2)2-2(x2+x-2)(2x+1) (-x2 -6x -5)] / (x2+x-2)4
The second derivative test at x=-5, gives a +ve value indicating that the gradient is increasing at x=-5. This suggests that x=-5 is actually a minimum point.
Why do I find different answers? If it’s an inflection point shouldn’t the second derivative equal 0 at x=-5, and if it’s a minimum point as the second derivative test suggests, shouldn’t dy/dx be –ve to the left of -5 but +ve to the right.
 
hndalama said:
Find the nature of the turning points? why are these two alternative methods giving me different answers.

Dy/dx= (-x2 -6x -5)/(x2+x-2)2
(-x2 -6x -5)/(x2+x-2)2 =0
-x2 -6x -5 =0
X2+6x+5=0
(x+5)(x+1) = 0
So turning points at x = -1 or -5
Click to expand...
That ought to say, "possible turning points". It's not always true that a point where dy/dx equals zero is a turning point. (Consider y=x^3, as one example.)


First derivative test
Track gradient around critical points
The denominator is squared so the denominator will always be positive hence I only consider the numerator i.e. (-x2-6x-5).
X-2-10-5-6
Gradient sign+0-0-

at x=-5, the gradient stays –ve indicating that it is an inflection point.
Click to expand...
It looks like you're checking the derivative at the critical points, instead of checking it on each side of them. I'm also not sure why you're looking at -6, -2, and 0.

For x = -1, check the derivative on each side, like at -1.1 and -0.9 (if the derivative has the same sign on each side of -1, then -1 is not a turning point; if the derivative is negative to the left and positive to the right, then -1 is a local minimum; if the derivative is positive to the left and negative to the right, then -1 is a local maximum).

Check -5 the same way. What is the derivative on each side of -5?

By the way, -2 is not in the domain of either y or dy/dx, so I'm not sure how you got + in your chart there.


… shouldn’t dy/dx be –ve to the left of -5 but +ve to the right[?]
Click to expand...
Why do you think that dy/dx is not negative to the left of x=-5 (and not positive to the right)?

PS: Please do not use textspeak, like -ve. If you have strength for typing only three characters, type neg instead. Same goes for +ve (type pos). ;)
 
mmm4444bot said:
By the way, -2 is not in the domain of either y or dy/dx, so I'm not sure how you got + in your chart there.
Click to expand...

Oh yes, you're right. As I mentioned, I was only considering the values in the numerator, so that’s why I didn’t notice that -2 is not in the domain of dy/dx. However, the gradient sign is still positive when I use another value like -1.5.

It looks like you're checking the derivative at the critical points, instead of checking it on each side of them. I'm also not sure why you're looking at -6, -2, and 0.
Click to expand...

No, I included the critical points to distinguish them from the points on each side of them. That’s why I looked at -6, -2 and 0. These are the points on each side of the critical points that I chose.
Since the gradient can only change sign at the critical points, picking -1.1, -1.5 or even -4 will give me gradients with the same sign, so it is not necessary to pick a point that is very close to the critical point. it just has to be in between the critical points.

Looking at this again with fresh eyes, I’ve realized my mistake. It is actually quite simple… even silly.
If you look at the order in which I wrote the x values in the table, for some reason I put -5 and then -6 after 0 in that table. This is because last night my mind deluded itself into believing that -5 and -6 were +5 and +6 whilst still managing to write -5 and -6.
I thought the gradient sign was negative on both sides of -5 because that’s what it looks like in the table. However when I rearrange them in the correct order, i.e. -6, -5, -1.5, -1, 0 the calculations are actually correct and consistent.

aahh sorry guys, please forgive my clumsiness.:(
 
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