The points of intersection with the X-axis coordinate
Graph of the function intersects the axis X at f = 0 so we need to solve the equation: acos(x2+12x)=0 Solve this equation The points of intersection with the axis X:
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0: substitute x = 0 to acos((2*x)/(1 + x^2)). acos(02+10⋅2) The result: f(0)=2π The point:
(0, pi/2)
Extrema of the function
In order to find the extrema, we need to solve the equation dxdf(x)=0 (the derivative equals zero), and the roots of this equation are the extrema of this function: dxdf(x)= the first derivative −−(x2+1)24x2+1−(x2+1)24x2+x2+12=0 Solve this equation Solutions are not found, function may have no extrema
Inflection points
Let's find the inflection points, we'll need to solve the equation for this dx2d2f(x)=0 (the second derivative equals zero), the roots of this equation will be the inflection points for the specified function graph: dx2d2f(x)= the second derivative −(x2+1)2−(x2+1)24x2+14xx2+14x2−3+(x2+1)(−(x2+1)24x2+1)2(x2+12x2−1)2=0 Solve this equation The roots of this equation x1=0
Сonvexity and concavity intervals: Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points: Concave at the intervals [0,∞) Convex at the intervals (−∞,0]
Even and odd functions
Let's check, whether the function even or odd by using relations f = f(-x) и f = -f(-x). So, check: acos(x2+12x)=acos(−x2+12x) - No acos(x2+12x)=−acos(−x2+12x) - No so, the function not is neither even, nor odd