How can we show that a continuous function satisfies a specific inequality?

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In summary, a continuous function is a type of mathematical function that is "smooth" and does not have sudden breaks or jumps. To prove that a function is continuous, we must show that it satisfies the three conditions of continuity. An inequality is a mathematical statement that compares two values or expressions. To show that a function satisfies a specific inequality, various methods such as graphing, algebraic manipulation, or calculus techniques can be used. Proving that a continuous function satisfies a specific inequality is important in understanding and analyzing the behavior of the function in real-world applications.
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Ackbach
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Here is this week's POTW:

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Let $f : [a,\infty)\to \Bbb R$ be a continuous function that satisfies the inequality $\displaystyle f(x) \le A + B\int_a^x f(t)\, dt$, where $A$ and $B$ are constants with $B < 0$. If $\displaystyle \int_a^\infty f(x)\, dx$ exists, show that $\displaystyle \int_a^\infty f(x)\, dx \le -A/B$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to Opalg for his correct solution.

[sp]Let \(\displaystyle F(x) = \int_a^x\!\!\! f(t)\,dt\) and let \(\displaystyle L = \int_a^\infty\!\!\! f(t)\,dt\). Then $F(x) \to L$ as $x\to\infty.$

Since $F$ is differentiable, with derivative $f$, it follows from the mean value theorem that (for each integer $n \geqslant a$) $F(n+1) - F(n) = f(x_n)$ for some $x_n \in (n,n+1).$ But $F(n+1) - F(n) \to L-L = 0$ as $n\to\infty$. Therefore $f(x_n) \to0.$

Now let $n\to\infty$ in the inequality \(\displaystyle f(x_n) \leqslant A + B\int_a^{x_n}\!\!\!f(t)\,dt\) to get $0\leqslant A + BL$. This inequality gets flipped when we divide though by the negative number $B$, giving $0 \geqslant \frac AB + L$, or $L\leqslant -\frac AB.$

Ackbach Comment: Opalg avoids the problem of $f(x)\not\to 0$ as $x\to\infty$ by choosing a subsequence $\{x_n\}\to \infty$ and $\{f(x_n)\}\to 0$ based on the Mean Value Theorem. Very clever!
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FAQ: How can we show that a continuous function satisfies a specific inequality?

What is a continuous function?

A continuous function is a type of mathematical function that has the property of being "smooth" or without any sudden breaks or jumps. This means that as the input values of the function change, the output values also change in a smooth and continuous manner.

How can we prove that a function is continuous?

To prove that a function is continuous, we must show that it satisfies the three conditions of continuity: the function exists at a given point, the limit of the function at that point exists, and the value of the function at that point is equal to the limit. This can be done using various methods such as the epsilon-delta definition, the intermediate value theorem, or the continuity theorem.

What is an inequality?

An inequality is a mathematical statement that compares two values or expressions using the symbols <, >, ≤, ≥, or ≠. It indicates that one value is less than, greater than, less than or equal to, greater than or equal to, or not equal to another value.

How can we show that a function satisfies a specific inequality?

To show that a function satisfies a specific inequality, we can use various methods such as graphing, algebraic manipulation, or calculus techniques such as differentiation and integration. These methods can help us determine the intervals where the function satisfies the inequality.

Why is it important to prove that a continuous function satisfies a specific inequality?

Proving that a continuous function satisfies a specific inequality is important because it allows us to understand and analyze the behavior of the function over a given interval. This information is useful in various fields such as economics, physics, and engineering where continuous functions are used to model real-world phenomena and make predictions about their behavior.

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