Graphing a function using calculus

Sniperman724
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How would you graph the function:
(x2-9)/(x2-4)

I am having a lot of trouble finding the critical and hypercritical numbers without the aid of a graphing utility.
Thank you very much!
 
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Sniperman724 said:
How would you graph the function:
(x2-9)/(x2-4)

I am having a lot of trouble finding the critical and hypercritical numbers without the aid of a graphing utility.
Thank you very much!

Start by dividing out the polynomial to give you something like:

y=f(x) + (something)/(x2-4)

Then factor (x2-4) and see what happens as x→±2.
 
To get the critical numbers, find the derivative of (x^2 - 9)/(x^2 - 4). The critical numbers are those for which the derivative is zero.
 
yea, i understand that, but I calculated the second derivative to be (-30x2-40)/(x2-4)3, which has two asymptotes at 2,-2, but the top of this fraction can never equal zero and I end up with imaginary numbers. How do you solve for concavity with imaginary numbers?
 
Where is the first derivative equal to zero? You seem to have skipped right over that step.

I got the same as you for the second derivative, so that would suggest that we're both right. The numerator is always negative for any real number, but the denominator can be positive or negative, depending on whether |x| > 2 or |x| < 2, respectively. That tells you the intervals where the graph of the function is concave up or down.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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