# Knowing the existence of max/min values of function in interval without taking derivative?

I'm given the following question.

"How do we know that the following function has a maximum value and a minimum value in the interval $$[0,3]$$"

$$f(x)=\frac{x}{x^2+1}$$

Is it possible to understand that there is a maximum and minimum value in the interval without taking the derivative of the function?

Is the solution to take the derivative of $$f(x)$$ and equate the resulting function to zero and solve?

• Have you heard of the extreme value theorem? – Theo Diamantakis Aug 20 '19 at 14:26
• indeed I have but have no experience applying it which is probably the reason for the question :) I will review the theorem and see how it can be applied. – esc1234 Aug 20 '19 at 14:29
• of course. Upon reviewing the theorem it clearly states that a continuous function on a closed interval will have a minimum and maximum value. Thanks very much! – esc1234 Aug 20 '19 at 14:36
• Do you have an intuitive idea why that must be so, though? ANd why you might not be able to state the same thing for $x \in (0,3)$? – fleablood Aug 20 '19 at 15:38
• I believe so. Open interval of the problem of always being able to get closer and closer to the endpoint therefore it's technically impossible to get a max or min at an endpoint. Hence the closed interval stipulation. I get it now but I'm just used to applying these theorems. Thanks though – esc1234 Aug 20 '19 at 15:39

"The extreme value theorem applies as $$\frac {x}{x^2 +1}$$ is continuous on the closed interval $$[0,3]$$"

The question asks you to show it has a max and min on that interval; not to find them. ANd the question doesn't ask about local maxima/minima but global extrema of all values on the interval.

The extreme value theorem says, and I quote: if a real-valued function $$f$$ is continuous on the closed interval $$[a,b]$$, then f must attain a maximum and a minimum, each at least once.

And $$\frac {x}{x^2+1}$$ is continuous on $$[0,3]$$. So the extreme value theorem says it attains a max and minimum at least once on the interval.

so that's it. Done.

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The intuitive idea for me is that the graph of a continuous function has a distinct starting point at $$[a,f(a)]$$ and a distinct ending point at $$[b,f(b)]$$ and for every point in between will have some actual real value. It's continuous so it can't "stretch" in any unbounded infinite value. So somewhere in there, either at one of the endpoints or somewhere between, it will get as big as it gets (within that interval, it can get bigger in other places outside the interval) and somewhere it will get as small as it gets.

That argument is naive and applies to ill-defined ideas (the most abusive is the it can't "stretch" to infinity) and needs to be formally defined and proven. And that is exactly what the Extreme Value theorem states.

In this case we have a path that starts at the point $$(0,f(0) = 0)$$ and ends at the point $$(3, f(3) = \frac 3{10})$$. In between in follows some curvy path between those two points. Somewhere there must be some point that was the biggest $$f(x)$$ value, and some that must be the least $$f(x)$$ value.

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We can get more specific. $$x \ge 0$$ and $$x^2 + 1 \ge 1$$ so $$f(x) \ge \frac x{x^2 + 1} \ge 0$$ so $$f(0) = 0$$ is a minimum. $$x \le 3$$ and $$x^2 + 1 \ge 1$$ so $$f(x)\le \frac 31$$ and so all the possible $$f(x)$$ are bounded above by $$3$$.

It is the definition of real numbers that if a set is bounded above that a least upper bound exist and a consequence of $$f$$ being continuous means that there is an $$x=c$$ where $$f(c) = \sup \{f(x)|x \in [a,b]\}$$.

We can go further an not $$f(1) = \frac 12 > \frac 3{10} =f(3)$$ so the max is not $$x=3$$ but some $$x: 0< x < 3$$ and so the maximum will be a local max and we COULD find it by taking the derivative.

But the question isn't ASKING us to. The question is only asking us to argue that there is a max. An the only thing we have to say for that is:

"The extreme value theorem applies as $$\frac {x}{x^2 +1}$$ is continuous on the closed interval $$[0,3]$$"

• Exactly. This is the correct answer. Thank you – esc1234 Aug 20 '19 at 15:45

Use that $$\frac{1}{2}\times\frac{2x}{x^2+1}\le \frac{1}{2}$$ and $$\frac{-1}{2}\le \frac{2x}{x^2+1}\times \frac{1}{2}$$

• Thanks. Has this idea come from the extreme value theorem which was suggested in the comment above? – esc1234 Aug 20 '19 at 14:30
• No, it comes from the AM-GM inequality. – Dr. Sonnhard Graubner Aug 20 '19 at 14:31
• Okay, the AM-GM inequality is something I haven't come across in my coursework yet. I will study it but because it hasn't come up in my coursework yet I'm thinking there must be another method. – esc1234 Aug 20 '19 at 14:34
• It is $$\frac{a^2+b^2}{2}\geq ab$$ for $$a,b\geq 0$$ – Dr. Sonnhard Graubner Aug 20 '19 at 14:36
• This shows that all values of $f(x): 0 \le x \le 3$ are so that $-\frac 12 \le f(x) \le \frac 12$ but not that there is a specific $c,d$ so that $-\frac 12 \le f(c) \le f(x) \le f(d) \le \frac 12$. – fleablood Aug 20 '19 at 15:34

The question gives a function that is continuous over a closed interval. The extreme value theorem states that a function that is continuous on a closed interval will have both a minimum and maximum value on that interval. The question requires that the student knows and understands the extreme value theorem.

We show that your function has a turning point in $$[0,3]$$ which is a maximum.

Note that $$\frac {x_1}{x_1^2+1}=\frac {x_2}{x_2^2+1}$$ for values where $$x_1\ne x_2$$ and $$x_1x_2=1$$

For example $$\frac {2}{2^2+1}=\frac{1/2}{(1/2)^2 +1}$$

Thus you have a turning point between $$1/2$$ and $$2$$.

Thus the turning point happens at $$x=1$$ and it is a maximum because for example $$f(1)=1/2 > f(2)=2/5$$