Inflection point is the point where the second derivative of a graph is zero.
y = (x+1)arctan xy' = (x+1)(arctan x)' + (1)arctan xy' = (x+1)/(x^2+1) + arctan xy'' = (x+1)(1/(1+x^2))' + 1/(1+x^2) + 1/(1+x^2)y'' = (x+1)(-1/(1+x^2)^2)(2x)+2/(1+x^2)y'' = ((x+1)(-2x)+1+x^2)/(1+x^2)^2y'' = (-2x^2-2x+2+2x^2)/(1+x^2)^2y'' = (-2x+2)/(1+x^2)^2
Solving for point of inflection: y'' = 00 = (-2x+2)/(1+x^2)^20 = -2x+2x = 1y(1) = (1+1)arctan(1) = 2 * pi/4 = pi/2
Therefore, E(1, pi/2).
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
There are 4 terms. Those terms are 4n, 2n, 9, and 5.
Answer:
Point B is located where x is negative and y is positive.
Hope this helps!!
