Finding the min value using the derivative

In summary, the conversation discusses finding the minimum value of a given function using its derivative. The function has no value at x=1 and has a vertical asymptote at x=-1. The derivative is positive for x in the interval (-1,1), and the function is decreasing in this interval. The domain of the function is (-1,1], and the domain of the derivative is (-1,1) without the point x=1. The conversation also explores the conditions for finding the extremum, and how they are not met in this case, yet the minimum value still exists due to the function being decreasing on its domain.
  • #1
TheColector
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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.
 
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  • #2
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
 
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  • #3
TheColector said:
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.
TheColector said:
For x = - 1 there is a vertical asymptote and f(x) goes to + infinity.
That's not where the asymptote is.
TheColector said:
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.
 
  • #4
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
 
  • #5
TheColector said:
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##.
 
  • #6
BvU said:
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 ?
 
  • #7
Mark44 said:
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.
 
  • #8
TheColector said:
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?
 
  • #9
TheColector said:
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.
 
  • #10
Mark44 said:
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.
 
  • #11
TheColector said:
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].
TheColector said:
So I can expect the min value to be at the end point of the domain
Yes, namely at x = 1.
TheColector said:
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

TheColector said:
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?
 
  • #12
Ray Vickson said:
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.
 
  • #13
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
 
  • #14
TheColector said:
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)).

minimum.png
 

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  • #15
FactChecker said:
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.
 
  • #16
TheColector said:
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.
 

Related to Finding the min value using the derivative

1. How do you find the minimum value using the derivative?

To find the minimum value using the derivative, you need to first find the critical points of the function by setting the derivative equal to zero. Then, use the second derivative test to determine if the critical point is a minimum or not. If the second derivative is positive, the critical point is a minimum.

2. Why is finding the minimum value using the derivative important?

Finding the minimum value using the derivative is important because it allows us to find the lowest point on a curve, which can be useful in many real-world applications. It also helps us understand the behavior of a function and make predictions about its behavior in the future.

3. Can the minimum value of a function be found without using the derivative?

Yes, the minimum value of a function can be found without using the derivative by graphing the function and visually identifying the lowest point on the curve. However, this method may not be accurate and can be time-consuming for more complex functions. Using the derivative provides a more precise and efficient way to find the minimum value.

4. Is the minimum value always a critical point?

No, the minimum value is not always a critical point. It is only a critical point if the second derivative is positive. If the second derivative is negative, the critical point is a maximum, and if the second derivative is zero, the critical point could be a minimum, maximum, or neither (a point of inflection).

5. Can the minimum value of a function change?

Yes, the minimum value of a function can change if the function is not continuous. For continuous functions, the minimum value will remain the same unless there is a change in the function's equation. Also, if the domain of the function changes, the minimum value can also change.

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