show the process to get the minimum and maximums, and critical points of y= 2x / (x^2 + 1)?

Answer 1

y= 2x / (x^2 + 1)
y' = [2(x^2+1)-2x(2x)]/(x^2+1)^2 = (2-2x^2)/(x^2+1)
Setting this = 0 we find x = +/- 1 are critical points.
When x = -1, f(x)= -2/2= -1
When x = 1 f(x) = 2/2 = 1
So max at x = +1 and min at x = -1
f(-x) = -f(-x) so f(x) is odd and has symmetry w/r to origin.
Lim f(x) x --> infinity = 0 so the x-axis is a horizontal asymptote.
f(x) = 0 when x=0, so origin is bothe x- and y-intercept.

Answer 2

y= 2x / (x^2 + 1)
=> y (x^2 + 1) = 2x
=> (x^2 + 1)*dy/dx + 2xy = 2 ... ( 1 )
For extreme values of y, dy/dx = 0
=> 2xy = 2
=> x * 2x / (x^2 + 1) = 1
=> 2x^2 = x^2 + 1
=> x^2 = 1
=> x = 1 or -1

For maximum value of y, dy/dx = 0 and d2y/dx2 < 0
Differentiating eqn. ( 1 ) w.r.t. x,
2x * dy/dx + (x^2 + 1)*d2y/dx2 + 2y + 2x*dy/dx = 0
Putting dy/dx = 0
=> (x^2 + 1)*d2y/dx2 + 2y = 0
=> d2y/dx2 = - 2y / (x^2 + 1) = - 4x / (x^2 + 1)^2
d2y/dx2 < 0 for x = 1 and > 0 for x = -1

So, y is maximum for x = 1 and minimum for x = -1
Max. y = 1 (putting x = 1) and
Mi. y = -1 (putting x = -1)

For critical point, dy/dx = 0 and d2y/dx2 = 0.
There is no critical point.