<p>I just took a huge calc test yesterday and I think I got the correct answers but was hoping somebody could just look over my work (I wrote it out below) and check if it's right. The test was mainly just a bunch of differentiation applications. </p>
<p>For critical values, I came up with some results that didn't look quite right, (because the graph is a parabola shape), but I think maybe they were just local min/max instead of global?</p>
<p>I think the original equation was: f (x) = x^4 - 2x^2 + 13
So:
f ' (x) = 4x^3 - 4x = 4(x^3 -x )
0 = 4 (x^3 - x) --> 0 = x^3 - x --> x = 0 and +/- 1
f " (x) = 12x^2 - 4
f " (-1) = 12(-1^2) - 4 --> 16 = positive = local min
f " (0) = 12(0^2) - 4 --> -4 = negative = local max
f " (1) = 12(1^2) - 4 --> 16 = positive = local min</p>
<p>Does this look okay to all you Calc geniuses? </p>
<p>lol, yeah...that's what made me think that I had the wrong answer though...when I graphed it, it was in the shape of a parabola. So, I spent like 5 more minutes looking at my work, and then when I made a table I found that it increased a tiny bit and then decreased, so it had sort of little "trick extremas" that you wouldn't be able to see from just graphing it... I don't know, I guess I'm just a little paranoid because I want to keep my average above 100% in that class and this was the problem that really stumped me. Thanks again for the help!</p>
<p>Why are you finding mins and maxes in the second derivitive?
Mins and maxes are part of the first derivitive.
Concavity and points of inflection are 2nd derivitive....</p>