Draw graph given f'(x), f''(x), domain, etc. Please check?

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1. I was asked to roughly sketch a graph given the information seen, in the below picture, to the right of the graph.

28rcitw.jpg


2. My actual graph is done by hand and the lines are not as straight, but the general shape is the same.

Does it look approximately right?

The areas I am worried about are:
a) when it specifies being concave of convex between certain intervals: whether it means completely visible between those intervals as concave/convex or just beginning to appear concave/convex

b) where it says "no limit of y x -> -2" exists, whether or not I should have drawn it the same angle as I did, or started the one to the left of -2 from a different angle.
 
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First, label the axis, and adopt the dot convention where a solid dot contains that point and a circle indicates a hole. Everything looks okay until you get to the second derivitive/concavity. The points where the second derivative would be zero (in between the domain of concavity given) should be points, and after that it should have a curve. For instance, when it says that from 5 to 10 (for example) f'' is positive, that means the graph should "open upwards" from 5 to 10, and if f'' is negative from 10 to infinity, then it should curve back downwards.
 
Thanks, QuarkCharmer.

To you or anyone who sees, I'm just wondering if this one is more accurate? Again, aside from the straight line, which I know means f"(x) = 0.

2cd9hz.jpg
 
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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