Graphing a function using calculus

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Homework Help Overview

The discussion revolves around graphing the function (x²-9)/(x²-4), focusing on identifying critical and hypercritical numbers through calculus techniques. Participants express challenges in analyzing the function without graphing utilities.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss finding critical numbers by calculating the derivative and question the implications of imaginary numbers when analyzing concavity. There are suggestions to factor the denominator and explore behavior as x approaches ±2.

Discussion Status

The conversation is ongoing, with participants sharing their calculations and interpretations of the derivatives. Some guidance has been offered regarding the analysis of concavity and critical points, but there is no explicit consensus on the approach to handling imaginary numbers in this context.

Contextual Notes

Participants are working under the constraint of not using graphing utilities, which may limit their ability to visualize the function's behavior effectively.

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.
 

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