How Do You Determine Where a Function is Concave Up or Down?

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SUMMARY

The discussion focuses on determining the concavity of the function y=(4x)/(x^2+1) by analyzing its second derivative, y'' = (8x^5 - 16x^3 - 24x)/(x^2 + 1)^4. The inflection points are found by setting y'' = 0, leading to x = 0, x = ±sqrt(3), while ignoring the imaginary solution x = sqrt(-1). The intervals of concavity are confirmed as concave up on (-sqrt(3), 0) and (sqrt(3), infinity), and concave down on (-infinity, -sqrt(3)) and (0, sqrt(3)).

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  • Understanding of calculus concepts, specifically second derivatives
  • Familiarity with the quotient rule for differentiation
  • Knowledge of inflection points and their significance in graph analysis
  • Ability to use sign charts for determining intervals of concavity
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  • Review the application of the quotient rule in calculus
  • Study the concept of inflection points in greater detail
  • Learn how to construct and interpret sign charts for functions
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Students and educators in calculus, mathematicians analyzing function behavior, and anyone interested in advanced graphing techniques.

donjt81
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problem: Use the graphing strategy to sketch the graph of y=(4x)/(x^2+1). check the intervals where it is concave up and where it is concave down. Then graph it. please use sign charts.

to find this we have to first find y''.
so I used the quotient rule twice to get this
y'' = (8x^5 - 16x^3 - 24x)/(x^2 + 1)^4
to find the inflection points we set y'' = 0 and solve for x.
I have a question on this
while solving y'' = (8x^5 - 16x^3 - 24x)/(x^2 + 1)^4 = 0
i come across this step
(8x)(x^2 + 1)(x^2 - 3) = 0
so that would mean

8x = 0 this mean x = 0
(x^2 + 1) = 0 this means x = sqrt(-1)
(x^2 - 3) = 0 this means x = +-sqrt(3)

but do we just ignore the x = sqrt(-1) and conclude that the inflection points are x = 0, x = sqrt(3) and x = -sqrt(3)

So once we get the inflection points we use the sign charts to find concave up and concave down.

intervals where graph is concave up: (-sqrt(3), 0) & (sqrt(3), infinity)
intervals where graph is concave down: (-infinity, -sqrt(3)) & (0, sqrt(3))

are these intervals i found correct?
 
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i did not get the imaginary answer you got

i got this for the second derivative
using mathematica
-\frac{16x}{(x^2+1)^2} +4x\left(\frac{8x^3}{(1+x^2}^3} - \frac{2}{(1+x^2)^3} = 0

it looks like you are assuming the denominator to be zero... taht can't be. that would not make this expression equal to zero.
 

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