What are the coordinates of the inflection point on the graph of y=(x+1)arctanx
a. (-1,0)
b. (0,0)
c. (0,1)
d. (1, pi/4)
e. (1, pi/2)

work plzz : )

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).

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No because if you do the higher number minus the lower number you will get a positive. if you do it the other way around you will get a negative number

The answer is “faster”.