Given \(f\) is continous function at every real numbers, prove that \(f\) does have global maxima if given condition:

\[\lim_{x \to \infty} f(x) = \lim_{x \to -\infty} f(x) = -\infty\]

No vote yet

4 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestUse Rolle's Theorem(Or Lagrange's Mean Value Theorem as a general case) ..http://en.wikipedia.org/wiki/Rolle's_theorem..

Using it we see that \(f'(c)=0\) for some real \(c\).

We cannot have a global minima for \(f\) as \(-\infty\) is the least "value" that any function can "attain". So \(f\) has at least one global maximum.

E.g.

\(\rightarrow\)Look at the graph of \(y=-x^{2}\).. It satisfies the given conditions and has a global maxima at x=0.

Log in to reply