Calculus Problem Using Second Derivative Test
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Calculus Problem Using Second Derivative Test

[From: ] [author: ] [Date: 11-05-01] [Hit: ]
1, -1.From there I found the second derivative, 4(3x²-1).I then took the critical points I found from the first derivative and put them on a number line to test on intervals with the second derivative. More specifically,......
In my final exam review packet for my college calculus class, one of the problems is really giving me trouble.

The directions for the problem are: "Find all relative minimums and maximums for y= x^4 -2x^2+1 (use second derivative test)

The first thing I did was to find the first derivative, which was 4x(x²-1). Once you set that equal to zero you get critical points/numbers of 0, 1, -1.

From there I found the second derivative, 4(3x²-1).

I then took the critical points I found from the first derivative and put them on a number line to test on intervals with the second derivative. More specifically, I found test points between the critical numbers and tested those test points in the 2nd derivative function. Through that I found a relative Max at (-1,0) and a relative Min at (1,0).

I thought that was the answer, but the guide said different. It said the answer was: Rel. max (0,1) and rel. Min at (1,0) and (-1,0).

On a hunch, I tried another method and actually got the right answer. However, the method I used didn't include the second derivative at all.

What I did second was find the critical points just as I did before (0,1,-1), put them on a number line including test points just as before. But instead of testing using the second derivative, I used the first derivative function. Through that I got the same answer as in the answer guide.

That would be good, but the directions stated that I was to use the Second Derivative Test. What am I doing wrong?

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Your method eas perfect.
did you test and got answers as below ?
at x=0 y''=-4
at x=-1 y''=8
at x=1 y''=8
when value of y'' is -ve there is a maximum and when y'' is positive there is a minimum

hence at x=0 maximum and at x=-1 and x=1 minimum

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Did you check at x = -1 ? The first derivative there is 0 and, as you know, and the second derivative is the same as x = 1. That seems like an ideal candidate.

Don't just test between the critical points, test on each side of the critical points.

See these page for more information::

http://www.wolframalpha.com/input/?i=+y%…

http://www.wolframalpha.com/input/?i=+pl…

http://www.wolframalpha.com/input/?i=+pl…
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