212k views
0 votes
At which value of x is the slope of the curve y=(1+x^2)^-1 the steepest

1 Answer

7 votes
Since the slope at a certain point is the derivative, and by memory (1+x^2)^-1 =arctan(x), we just have to find arctan(x)'s highest point. Graphing it out based off of tan(x), we know that when x approaches π/2 in tan(x) then tan(x) approaches infinity, arctan(x) approaches π/2 when x approaches infinity. As arctan(x) gets closer to π/2 similar to how x gets closer to π/2 in tan(x) as well as that it gets a tiny bit higher (due to that it starts low) as x approaches infinity, your answer is infinity
User Kikiwa
by
7.2k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.