Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

I'm not sure what I did. Did I find the inflection point?

Answer 4

my bad, i didnt notice the 'what value of x' part....u r right, it's 0.

Answer 5

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.

Answer 6

ohhh....thanks!

Answer 7

Glad to help . :)