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Nanoman
Group Title
Check my work:
At what value of x does the function 1/(1+x^2) change from increasing to decreasing?
I'm not sure what I did, but I think it's right.
My work: Derivative 1/(1+x^2) > 2x/(1+x^2)^2 = 0. Solving for x gives 0. Therefore, the answer is zero.
 one year ago
 one year ago
Nanoman Group Title
Check my work: At what value of x does the function 1/(1+x^2) change from increasing to decreasing? I'm not sure what I did, but I think it's right. My work: Derivative 1/(1+x^2) > 2x/(1+x^2)^2 = 0. Solving for x gives 0. Therefore, the answer is zero.
 one year ago
 one year ago

This Question is Closed

tanvirms Group TitleBest ResponseYou've already chosen the best response.0
the answer is 1....x^2 is always positive, so 1+x^2 will have a minimum value of 0 (which is unacceptable as u cant divide anything by 0)....again, 1/1+x^2 will have maximum value 1 when x=0, so, the graph will ascend upto 1 and then start descending.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
Though your answer is right, but still you'll have to confirm if that is maxima ( goes from increasing to decreasing) For that, you'll have to find the 2nd derivative and check if it is positive or negative for x=0 If it is negative, then it is a maxima, otherwise not. 2nd derivative >(6x^2 2)/(1 + x^2)^3 for x=0 , it is negative,which justifies your answer.
 one year ago

Nanoman Group TitleBest ResponseYou've already chosen the best response.0
I'm not sure what I did. Did I find the inflection point?
 one year ago

tanvirms Group TitleBest ResponseYou've already chosen the best response.0
my bad, i didnt notice the 'what value of x' part....u r right, it's 0.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
That'd have been an inflection point only if second derivative of f(x) had come equal to 0 for x=0 but 2nd derivative is 2 i.e. negative here, hence it is not an infection point but is a maxima, where your function goes from increasing to deceasing.
 one year ago

Nanoman Group TitleBest ResponseYou've already chosen the best response.0
ohhh....thanks!
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
Glad to help . :)
 one year ago
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