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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
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

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tanvirmsBest 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

shubhamsrgBest 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

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

tanvirmsBest 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

shubhamsrgBest 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
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