Finding the min value using the derivative

[TheColector]

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.

 

[BvU]

Did you make a sketch of the function ?
[edit] don't erase the template. Show your working in detail
$f(x)$ has no value for $x=1$ (asymptote)
But it does have a value for $x=-1$
$f'(x)$ is > 0, not < 0

 

[Mark44]

First derivative has no solutions and it is < 0 for {-1 < x < 1} when f(x) is given for {-1 < x <= 1}.

No, f'(x) > 0 for x in that inteval, as BvU already mentioned.

For x = - 1 there is a vertical asymptote and f(x) goes to + infinity.

That's not where the asymptote is.

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?

An extremum can occur at three places:

1. A point at which f'(x) = 0
2. A point in the domain of f at which f'(x) is not defined, such as the point (0, 0) for f(x) = |x|.
3. An endpoint of the domain of f.

 

[TheColector]

Oh gosh
Sorry for that mistake. I entered wrong function. It should be sqrt((1-x)/(1+x))
The statements I posted describe function above.
Sorry for that stupid mistake of mine

 

[Mark44]

Oh gosh
Sorry for that mistake. I entered wrong function. It should be sqrt((1-x)/(1+x))
The statements I posted describe function above.
Sorry for that stupid mistake of mine

OK, no problem.
An important point to be considered is the domain of the function f. I agree that the equation f'(x) = 0 has no real roots, so the 2nd and 3rd points in my list in post #3 come into play. The domain of f is the set of points for which $\sqrt{\frac{1 -x }{1 + x}}$ is defined. To determine this, you need to solve the rational inequality $\frac{1 -x }{1 + x} \ge 0$.

 

[TheColector]

Did you make a sketch of the function ?
[edit] don't erase the template. Show your working in detail
$f(x)$ has no value for $x=1$ (asymptote)
But it does have a value for $x=-1$
$f'(x)$ is > 0, not < 0

Yes I did and there is obvious min of f(x) for x = 1, but how can this be that the conditions for finding the extremum of function are not met, yet there is the extremum at that point. Is it maybe because f(x) is not differentiable at point x = 1 ?

 

[TheColector]

OK, no problem.
An important point to be considered is the domain of the function f. I agree that the equation f'(x) = 0 has no real roots, so the 2nd and 3rd points in my list in post #3 come into play. The domain of f is the set of points for which $\sqrt{\frac{1 -x }{1 + x}}$ is defined. To determine this, you need to solve the rational inequality $\frac{1 -x }{1 + x} \ge 0$.

Domain is {-1 < x <= 1} - considering f(x)
when considering the domain of f'(x) it is the same but without point of 1
I guess that means there is no derivative at that point, as it goes to (- infinity) there.

 

[Mark44]

Domain is {-1 < x <= 1} - considering f(x)
when considering the domain of f'(x) it is the same but without point of 1
I guess that means there is no derivative at that point, as it goes to (- infinity) there.

So, since f' < 0 on the interval (-1, 1), can you say something about the maximum value of f and the minimum value of f?

 

[Ray Vickson]

Yes I did and there is obvious min of f(x) for x = 1, but how can this be that the conditions for finding the extremum of function are not met, yet there is the extremum at that point. Is it maybe because f(x) is not differentiable at point x = 1 ?

For a (differentiable) function $f(x)$on a bounded interval $[a,b]$ the derivative need not equal zero at the max or the min. The derivative should equal zero at interior optima, but that condition may fail when the solution is at an endpoint. For example, on the interval $[0,1]$ the solution of $\min x$ is at $x = 0$ and the solution of $\max x$ is at $x = 1$. The derivative is positive at both of these solutions.

 

[TheColector]

So, since f' < 0 on the interval (-1, 1), can you say something about the maximum value of f and the minimum value of f?

I can say that f(x) is decreasing in the domain of f'(x) as it is < 0. So I can expect the min value to be at the end point of the domain but I don't exactly know why the min value exists since the conditions for them to be are not met. That is what I have problem with, not exactly finding min value but finding the extremum at all.

 

[Mark44]

I can say that f(x) is decreasing in the domain of f'(x) as it is < 0.

f is decreasing on its domain, (-1, 1].

So I can expect the min value to be at the end point of the domain

Yes, namely at x = 1.

but I don't exactly know why the min value exists since the conditions for them to be are not met.

What does this mean? Going through the list I wrote
Is f'(1) = 0? No.f' is not defined at x = 1
Is f(1) defined but f'(1) not defined? Yes
Is 1 an endpoint of the interval in question? Yes

That is what I have problem with, not exactly finding min value but finding the extremum at all.

The min. value is one of the extrema. The max. value is the other extremum. Are you having a problem understanding what the words mean?

 

[Mark44]

For example, on the interval $[0,1]$ the solution of $\min x$ is at $x = 0$ and the solution of $\max x$ is at $x = 1$. The derivative is positive at both of these solutions.

No, since the OP made a correction to the function of this problem in post #4. For the revised function of this problem, on the interval [0, 1] (which is not the interval we're concerned with in this problem), the max value occurs at x = 0, and the min. value occurs at x = 1.

 

[TheColector]

I generally know the meaning of extremum and that is either min or max depending on a change of sign, the thing I HAD problem with was to show existence of the extremum using derivative only without wider analysis (such as decreasing value of function). I was pretty confused about all of it after collecting exam results from my prof as his demands are sometimes not quite logical. I get it now and thanks for your help and commitment.
Cheers m8s

 

[FactChecker]

I can say that f(x) is decreasing in the domain of f'(x) as it is < 0. So I can expect the min value to be at the end point of the domain but I don't exactly know why the min value exists since the conditions for them to be are not met.

You should not think of the derivative as being a condition for a minimum, but rather a symptom of a certain type of minimum. There are a few different ways that a function can have a minimum. When one is asked for a global minimum, he must first get a general idea of the global behavior of the function. It can have a minimum that it achieves as your function does (function f(x) below), or it can have an asymptotic limit as x goes to -∞ (function g(x)), or it can have a local minimum where the derivatives indicate a minimum that is not a global minimum (function h(x)).

 

[TheColector]

You should not think of the derivative as being a condition for a minimum, but rather a symptom of a certain type of minimum. There are a few different ways that a function can have a minimum. When one is asked for a global minimum, he must first get a general idea of the global behavior of the function. It can have a minimum that it does not achieve as your function does (function f(x) below), or it can have an asymptotic limit as x goes to -∞ (function g(x)), or it can have a local minimum where the derivatives indicate a minimum that is not a global minimum (function h(x)).

View attachment 219701

Thanks a lot
That is a nice way of putting it.

 

[Ray Vickson]

I can say that f(x) is decreasing in the domain of f'(x) as it is < 0. So I can expect the min value to be at the end point of the domain but I don't exactly know why the min value exists since the conditions for them to be are not met. That is what I have problem with, not exactly finding min value but finding the extremum at all.

You persist in making this error. In this example, $f(1)$ exists and really, truly IS the minimum. The fact that $f'(1)$ is $< 0$ just reinforces that fact, not contradict it!

You seem to be under the much mistaken impression that the derivative must be 0 at an optimum. NO, no, no! For an INTERIOR solution, that is, indeed, true; but on a bounded interval it is almost never true at an endpoint. That is why I introduced the simplest possible example to illustrate that fact, but I guess you did not read my previous response #9.