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This might seem like a very basic problem for the most of you but I am a little bit confused as to the problem of maximum and minimum values.

For, example I do understand why y = x^2 has no absolute maximum as y-->[itex]\infty[/itex] as x-->[itex]\infty[/itex]...

However, in the closed interval, say, [0, 2], does the function y = x^2 have a local maximum = 4 and a local minimum = 2 at it's endpoints? In this case it so happens that y'(0) = 0 and we have a local minimum there but would it have been the same for the closed interval [1, 2] (with min = 1 and max = 4)?

If someone asks you to motivate why a certain function has a maximum or minimum in a certain interval what would you say? In the latter of above examples, would it suffice to say, y is continuous and y>0 in our interval --> there be a number such that y(m) ≤ y(x) ≤ y(M) for all x in our closed interval. These max and min occurs either in critical points, where y'(x) = 0, in singular points, where y'(x) is not defined or in endpoints. And thereafter do calculations to show that y'(x) = 0 for x = 0, not in our interval and since y'(x) is defined for all x this means that our max and min is found in the endpoints.

Of course the above example is a very simplified one, but I want to try and understand the idea.

What about for an open interval?

Let's take this function as an example:

f(x) = [itex]\frac{x}{(x-1)(x-4)}[/itex], motivate why the function must have a largest value in the interval 1<x<4 and find this value.

Would this answer be correct?

f(x) is continuous and |f(x)| > 0 (f(x) < 0) in the interval (1, 4). As we approach 1+ and 4- f(x) --> -[itex]\infty[/itex]. Therefore, f(x) will have a maximum value in the interval (1, 4) in either a point where f'(x) = 0 or where f'(x) is not defined. Since f'(x) = [itex]\frac{-x^2 + 4}{((x-1)(x-4))^2}[/itex] and is continuous in the interval (1, 4) we have no singular points where f'(x) is not defined. This means, that the maximum value will be found in a point where f'(x) = 0.

For f'(x) to be zero it is required that its numerator, -x^2 + 4 = 0 --> x = [itex]\pm[/itex]2 and since only 2 is in the interval we have f(2) = -1 which is the maximum value of f(x) in the open interval (1,4).

I'm trying to learn the idea behind the mathematics and not just to learn how to solve the problems. I'm in my first year of BSc in mathematics so try to not explain in a too advanced way.

I appreciate any help and corrections.

Sincerely,

Mr. Fest

PS. What I mostly want to understand is the answer to the following question: "How do I answer the question: Why must a given function f(x) have a smallest or largest value or both in a given closed or open interval?"

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# Maximum and minimum values

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