Asked by John
Determine over what interval(s) is the function y=4x-6arctan(x) is concave up/concave down
I took the second derivative and ended up with x=0. I do not believe I did this right. IF I happen to be right, wouldn't our answer just be undefined, rather than having intervals being concave up or down?
I took the second derivative and ended up with x=0. I do not believe I did this right. IF I happen to be right, wouldn't our answer just be undefined, rather than having intervals being concave up or down?
Answers
Answered by
Steve
If you got a single value for an interval, it doesn't look good for the kid.
y = 4x-6arctan(x)
y' = 4 - 6/(x^2+1) = (4x^2-1)/(x^2+1)
y" = 12x/(1+x^2)^2
You are correct with what you got. But, what do you have when y" = 0? An inflection point. That's where the concavity changes. So, now you have to remember that the sign of y" indicates concavity. Here,
y" < 0 for x<0
y" > 0 for x>0
Got it now? You can write your intervals.
Just to verify, check
http://www.wolframalpha.com/input/?i=+4x-6arctan%28x%29
y = 4x-6arctan(x)
y' = 4 - 6/(x^2+1) = (4x^2-1)/(x^2+1)
y" = 12x/(1+x^2)^2
You are correct with what you got. But, what do you have when y" = 0? An inflection point. That's where the concavity changes. So, now you have to remember that the sign of y" indicates concavity. Here,
y" < 0 for x<0
y" > 0 for x>0
Got it now? You can write your intervals.
Just to verify, check
http://www.wolframalpha.com/input/?i=+4x-6arctan%28x%29
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