Difference between revisions of "Trivial Inequality"

(created simple definition, the \ge command messes up the page, HELP!)
 
(The Inequality)
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The trivial inequality states that <math> x^2 >= 0 </math> for all x. This is a rather useful inequality for proving that certain quantities are non-negative. The inequality appears to be obvious and unimportant, but it can be a very powerful problem solving technique.
 
The trivial inequality states that <math> x^2 >= 0 </math> for all x. This is a rather useful inequality for proving that certain quantities are non-negative. The inequality appears to be obvious and unimportant, but it can be a very powerful problem solving technique.
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=== Applications ===
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'''Maximizing and minimizing quadratic functions'''
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Let <math>f(x)=ax^2+bx+c</math> be a function of degree two, that is, a quadratic function, where <math>a\in\{1,-1\}</math>. If <math>{a}=1</math>, then <math>{f}</math> has only a minimum; if <math>a=-1</math>, then <math>{f}</math> has only a maximum. We can 'complete the square' in this function as follows:
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<math>ax^2+bx+c = a\left(x+\frac{b}{2}\right)^2-a\cdot \frac{b^2}{4}+c</math>
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By the trivial inequality, the minimum/maximum is then easily determined [too lazy to explain].

Revision as of 16:28, 17 June 2006

The Inequality

The trivial inequality states that $x^2 >= 0$ for all x. This is a rather useful inequality for proving that certain quantities are non-negative. The inequality appears to be obvious and unimportant, but it can be a very powerful problem solving technique.

Applications

Maximizing and minimizing quadratic functions

Let $f(x)=ax^2+bx+c$ be a function of degree two, that is, a quadratic function, where $a\in\{1,-1\}$. If ${a}=1$, then ${f}$ has only a minimum; if $a=-1$, then ${f}$ has only a maximum. We can 'complete the square' in this function as follows:

$ax^2+bx+c = a\left(x+\frac{b}{2}\right)^2-a\cdot \frac{b^2}{4}+c$

By the trivial inequality, the minimum/maximum is then easily determined [too lazy to explain].