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

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To determine where the function y=(4x)/(x^2+1) is concave up or down, the second derivative y'' is calculated using the quotient rule, resulting in y'' = (8x^5 - 16x^3 - 24x)/(x^2 + 1)^4. Inflection points are found by setting y'' = 0, leading to the equation (8x)(x^2 + 1)(x^2 - 3) = 0, which gives x = 0, x = ±sqrt(3) as valid inflection points while ignoring the imaginary solution. The intervals of concavity are identified as concave up on (-sqrt(3), 0) and (sqrt(3), infinity), and concave down on (-infinity, -sqrt(3)) and (0, sqrt(3)). The discussion clarifies that the denominator cannot be zero, ensuring the second derivative remains valid. Understanding these intervals is crucial for accurately graphing the function's behavior.
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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