# Sketching a function given certain properties

The following is the last problem on a practice exam and it is giving me trouble.

Given the following information about a continuous function, $f$, how do I sketch a possible graph of $f$:

$f(-4)=3, f'(-4)$ does not exist ; $f(-2)=0, f'(-2)=0;~ f(0)=-4, f'(0)$ does not exist;
$f(2)=0, f'(2)=0; f(4)=3, f'(4)$ does not exist;
$f'(x) \lt 0$ when $x\lt -4, -4\lt x \lt -2, -2\lt x \lt 0, x\gt 4$;
$f'(x) \gt 0$ when $0\lt x \lt 2, 2\lt x \lt 4$;
$f''(x) \lt 0$ when $x\lt -4, -2\lt x\lt 0, 0\lt x \lt 2$;
$f''(x) \gt 0$ when $-4\lt x \lt -2, 2\lt x \lt 4, x\gt 4$.

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Get a piece of paper with squares, tick off your range of $x$ from -4 to 4.
In terms of the curve's looks, what do the different assertions about the function mean? What does "$f'(-4)$ doesn't exist" mean? What does "$f'(x) < 0$ if $x > 4$" mean? What does it mean that $f''(x) < 0$? The "doesn't exist" can mean different things ($f'(x)$ doesn't exist if the derivative has a discontinuity at $x$, which could mean that it tends to $\pm \infty$ or has a jump).
For a bonus, try to splice together a simple function that works as announced, by starting with a polynomial with the right overall form and multiplying by appropiate $\pm 1 / (x - x_0)$ to get points where it tends to $\pm \infty$, and tweak so it goes through the prescribed points. Then use some graphing software to hand in a neat sketch ;-)
Thanks. Ok. so I know what with the given conditions, $f$ is increasing or decreasing on certain intervals. $f$ is also concave up/down on certain intervals. What is troubling me is $f'$ does not exist. Will I be correct to interpret that to mean that there's a kink at those points? – Gorg Mar 27 '13 at 3:03