- #1

TheColector

- 29

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Member reminded that the HW template must be used

## Homework Statement

Hi

I'm having a trouble with finding min value of given function: f(x) = sqrt((1+x)/(1-x)) using derivative.First derivative has no solutions and it is < 0 for {-1 < x < 1} when f(x) is given for {-1 < x <= 1}.

For x = - 1 there is a vertical asymptote and f(x) goes to + infinity. Because f'(x) < 0 that means f(x) is decreasing from - 1 to 1.

How am I to find the extremum when both sufficient and necessary conditions are not met?

The sufficient condition for extremum tells us that function must be

**continuous**at a

**point**x=a(1 in this case) and there must be a change of sign for f'(x) (from - to + in order to be a min).

In this case there is no change of sign around point of x = 1 but function is continuous in that very point.

I can assume the min exists for x = - 1 just because function is decreaing to that very point and it has the value of 0 in this point.

My question is how to tell there is a min using derivative.