Question

Calculus,

Can you tell me how to solve this?
http://screencast.com/t/oLnAsIX9jdLr

Answer 1

you have to draw approximately the graph of each function by using the information such as : minimum, concavity, .. these things are easy to obtain using the derivative of the function.
For example, \(p(x)=x^2-3x-4\). After some calculus you get: \(p(x)=(x-4)(x+1)\). The roots are 4 and -1. This is useful for drawing.

Answer 2

Second step: take the derivative and equalize to zero: \(p'(x)=0\), which is here \(2x-3=0\). the solution is \(x=3/2\). Another known point \((\frac32, p(\frac32))\).
At this point (using the interpretation of the derivative) you know that there's a maximum or minimum. Looking at \(p''(3/2)\) will tell you (you have to know how to interpret it). After that you might be able to draw it.. good luck. af

Answer 3

So it's either: get a lot of points,
OR
Get the roots, the local maxima and local minima, and use \(p''\) to tell you the concavity in each of these points.

Answer 4

\(p(x)=1+8x-x^2\). then \(p'(x)=8-2x\).
The solution of \(p'(x)=0\) is \(x=4\). There's a max or min in 4. Check it with \(p''(4)\).
Then you know it looks like :