What is the Mean Value Theorem Inequality for the Interval [0,1]?

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Homework Help Overview

The discussion revolves around demonstrating an inequality involving the function \(\sqrt[3]{1+x}\) over the interval [0,1], specifically showing that \(\frac{1}{4}x+1\leq\sqrt[3]{1+x}\leq\frac{1}{3}x+1\). Participants are exploring the Mean Value Theorem in relation to this inequality.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to manipulate the inequality by subtracting 1 and dividing by \(x\), leading to a derivative analysis. Some participants question the validity of the inequality by testing specific values, while others discuss the concavity of the function and its relationship to tangent and secant lines.

Discussion Status

The discussion is active, with participants offering various insights and questioning the assumptions behind the inequality. There is no explicit consensus on the truth of the inequality, and multiple interpretations of the problem are being explored.

Contextual Notes

Some participants express uncertainty about the validity of the inequality, particularly when evaluating it at the endpoints of the interval. There is also mention of the need for a function with specific derivative constraints, indicating potential gaps in information or understanding.

mtayab1994
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Homework Statement



For every x in the interval [0,1] show that:j

[tex]\frac{1}{4}x+1\leq\sqrt[3]{1+x}\leq\frac{1}{3}x+1[/tex]

The Attempt at a Solution



Well i subtracted 1 from all sides and divided by x and I got:

[tex]\frac{1}{4}\leq\frac{\sqrt[3]{1+x}-1}{x}\leq\frac{1}{3}[/tex]

But now I need to find a function that has a derivative with a max value of 1/3 and a min value of 1/4 there where I'm stuck. Any help would be very much appreciated.
 
Last edited:
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Is the inequality even true? Let x = 0, you have -1 < 1 < -1
 
MarneMath said:
Is the inequality even true? Let x = 0, you have -1 < 1 < -1

Sorry my mistake i edited it.
 
Not sure quite how this works in with the mean value theorem but f(x)=(x+1)^(1/3) is concave down. x/3+1 is the tangent line at x=0, so it lies above f(x). x/4+1 is below the secant line connecting x=0 and x=1. So it lies below f(x).
 
Last edited:

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