For f(x)=x/x^2+1,s show that the slope m of the tangent line to y= f(x) satisfies -1/8≤m≤1?
1 Answer
We have:
#f(x) = x/(x^2 +1)#
Taking the derivative w.r.t
#f'(x) = (1(x^2 + 1) - x(2x))/(x^2 +1)^2#
#f'(x) =(x^2 + 1 - 2x^2)/(x^2 + 1)^2#
#f'(x) = (1 - x^2)/(x^2 + 1)^2#
We wish to find the maximum/minimum of this.
#f'(x) = (1- x^2)/(x^4+ 2x^2 + 1)#
#f''(x) = (-2x(x^4 + 2x^2 + 1) - (1 - x^2)(4x^3 + 4x))/(x^2 + 1)^4#
#f''(x) = (-2x^5 - 4x^3 - 2x - (4x^3 - 4x^5 + 4x - 4x^3))/(x^2 +1)^4#
#f''(x) = (2x^5 - 4x^3 -6x)/(x^2 +1)^4#
This has critical points when
#0 = 2x^5 - 4x^3 - 6x#
#0 = 2x(x^4 - 2x^2 - 3)#
#0 = 2x(x^2 - 3)(x^2 + 1)#
#x = 0 or +-sqrt(3)#
The slope of the tangent at
#f'(0) = (1 - 0^2)/(0^4 + 2(0)^2 + 1) = 1#
The slope of the tangent at
#f''(sqrt(3)) = f''(-sqrt(3)) = (1 - 3)/(4)^2 = -2/16 = -1/8#
Thus, the slopes of all the tangents must be in the interval
Hopefully this helps!