Question

What are the absolute extreme values of the function `f(x) = x/sqrt(x^2+1)` on the interval `[0,2]`?

Answer 1

This function is strictly increasing on the given interval, so the absolute minimum value is `f(0)=0` and the absolute maximum value is `f(2)=2/sqrt(5) approx 0.894427`.

Explanation:

By the Quotient Rule and Chain Rule, the derivative of this function is `f'(x)=(sqrt(x^2+1)-x*(1/2)(x^2+1)^(-1/2)*2x)/(x^2+1)`.

This can be simplified by multiplying the top and bottom by `sqrt(x^2+1)=(x^2+1)^(1/2)` to get

`f'(x)=(x^2+1-x^2)/(x^2+1)^(3/2)=1/(x^2+1)^(3/2)`.

From this, we see that `f'(x)>0` for all `x`. Therefore, `f` is strictly increasing on any interval, including the interval `[0,2]`. This implies that the absolute minimum value is `f(0)=0` and the absolute maximum value is `f(2)=2/sqrt(5) approx 0.894427`.