# What theorem is this?

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

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

Note by Nabila Nida Rafida
4 years, 9 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Use 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.

- 4 years, 9 months ago