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Andreyy89
3 years ago
10

I rlly need help :( this is hard

Mathematics
1 answer:
GrogVix [38]3 years ago
8 0

Answer:

A

Step-by-step explanation:

The domain does not exist when the denominater of an equation is zero. So in this case if x was -2 and we added 2 the denominator would be zero, and that value does not exist in the domain. Hope this helps!

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What are the coordinates of the inflection point on the graph of y=(x+1)arctanx
Dennis_Churaev [7]

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!

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