Discover the Truth Behind Relative Maximums: A Simple Max/Min Question

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When given a table listing x, x', and x'' which of the following are true?

if at critical #, second derivative is negative, # is a relative max
if at critical #, second derivative is positive, # is a relative max

Thanks!
 
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These are yes or no questions. Read your book!
 
sorry, I know it seems like I'm not trying.
I was reviewing notes for a test tomorrow and in one section I wrote the first one and then right below it I wrote the second one. If I weren't so pressed on time I would find it in my book but I'm really trying to get finished and get a good nights sleep. D:
 
Remember this. y=x^2. Second derivative=2. x=0 is a min. y=-x^2. Second derivative=(-2). x=0 is a max.
 
all right! thanks.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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