When is a function bounded using differentiation

[sara_87]

Homework Statement

how do i determine whether a function is bounded using differentiation

eg: f(x)=x/(2^x)

Homework Equations

The Attempt at a Solution

i know it has something to do with maximums and minimums but i can't figure out how to do it.

any help would be appreciated. thank you

 

[Dick]

You could look at the limits of the function as it approaches plus and minus infinity. If both exist and are finite, and if the function is defined and continuous for all x, then it is bounded.

 

[EnumaElish]

Homework Statement

how do i determine whether a function is bounded using differentiation

eg: f(x)=x/(2^x)

Homework Equations

The Attempt at a Solution

i know it has something to do with maximums and minimums but i can't figure out how to do it.

any help would be appreciated. thank you

You can use differentiation to investigate the behavior of f. Say, the function is f(x) = x/2^x on x > 0. Then f'(x) = 2^-x (1 - x Log[2]), which has roots 1/Log[2] and +infinity. At x = 1/Log[2], f''(x) = 2^-x Log[2] (-2 + x Log[2]) is < 0, so you have the maximum. Note that f(x) > 0 for x > 0 and f(0) = 0. As x --> +infinity, f(x) --> 0 from above; but f(0) = 0 so x = 0 is the minimum. Since you can "account for" both the maximum and the minimum, f is bounded on x > 0.

 

[sara_87]

thank you very much.
what if we have:
f(x)=(-2x^2)/(4x^2-1)
i know that it's not bounded but i don't know why

 

[quantumdude]

The graph of that function has two vertical asymptotes. Functions don't get much more "unbounded" than that!

 

[sara_87]

actually i think it is bounded because there's no value of x that would make that function greater than 1
or is there?

 

[quantumdude]

Sure there is. As I said, the graph of that function has 2 vertical asymptotes. You can find values of x for which the function blows up to infinity, and down to negative infinity.

Do you know what I mean when I say "vertical asymptote"?

 

[sara_87]

yes i do know what vertical assymptotes are.
umm but i still didnt understand what u meant. you can find values of x for which the function blows down to -ve infinity but not up.
?

 

[quantumdude]

The function certainly does blow up to positive infinity, as you approach -1/2 from the right and as you approach +1/2 from the left.

 

[sara_87]

oh thank u very much
that helps.
just one last question:
same question as before but with function:
sqrt(x)/1000

is it not bounded since n continues to increase to infinity?

 

[quantumdude]

You're right that it's not bounded (on [itex][0,\infty)[/itex] that is--we really should be specifying an interval when making these statements).

But what's "n"?

 

[HallsofIvy]

Now, I'm confused as to what function you are talking about. The original function was f(x)= x/(2^x) which is definitely bounded on [itex][0, \infty)[/itex]. It is bounded "above" but not bounded "below" so is not bounded. I don't see any asymptotes when I graph it.

 

[quantumdude]

Posts 1 through 3 pertain to f(x)=x/(2^x).
Posts 4 through 9 pertain to f(x)=(-2x^2)/(4x^2-1).
Posts 10 and 11 pertain to f(x)=sqrt(x)/1000.

 

[EnumaElish]

Posts 1 through 3 pertain to f(x)=x/(2^x).

Post 3 pertains to f(x)=x/(2^x) for x > 0.