Question

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?

Answer 1

We have:

`f(x) = x/(x^2 +1)`

Taking the derivative w.r.t `x`.

`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 `f''(x) = 0`.

`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 `x= 0` will be a maximum:

`f'(0) = (1 - 0^2)/(0^4 + 2(0)^2 + 1) = 1`

The slope of the tangent at `x = +- sqrt(3)` will be a minimum.

`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 `-1/8 ≤ m ≤ 1`, as required.

Hopefully this helps!