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

In summary, to prove that for every x in the interval [0,1], \frac{1}{4}x+1\leq\sqrt[3]{1+x}\leq\frac{1}{3}x+1 is true, the attempt at a solution involved subtracting 1 from all sides and dividing by x. To complete the proof, a function with a derivative that has a maximum value of \frac{1}{3} and a minimum value of \frac{1}{4} was needed. The function (x+1)^(1/3) was suggested as it is concave down and its tangent line at x=0, x/3+1, lies above it while the
  • #1
mtayab1994
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0

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.
 
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  • #2
Is the inequality even true? Let x = 0, you have -1 < 1 < -1
 
  • #3
MarneMath said:
Is the inequality even true? Let x = 0, you have -1 < 1 < -1

Sorry my mistake i edited it.
 
  • #4
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).
 
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Related to What is the Mean Value Theorem Inequality for the Interval [0,1]?

What is the Mean Value Theorem inequality?

The Mean Value Theorem inequality is a mathematical theorem that states that if a function is continuous on a closed interval and differentiable on the open interval, then there exists at least one point within the interval where the slope of the tangent line is equal to the average rate of change of the function.

What is the significance of the Mean Value Theorem inequality?

The Mean Value Theorem inequality is significant because it allows us to find a specific point on a function where the instantaneous rate of change is equal to the average rate of change. This can be useful in solving problems related to optimization, such as finding the maximum or minimum value of a function.

What is the difference between the Mean Value Theorem and the Mean Value Theorem inequality?

The Mean Value Theorem states that there exists at least one point where the derivative of a function is equal to the slope of the secant line connecting the endpoints of an interval. The Mean Value Theorem inequality, on the other hand, states that there exists at least one point where the derivative is less than or equal to the slope of the secant line.

How is the Mean Value Theorem inequality used in real life?

The Mean Value Theorem inequality can be used in real life to solve problems related to optimization, such as finding the maximum or minimum value of a function. It can also be used in economics to analyze the average rate of change of a variable over a given time period.

What are some common misconceptions about the Mean Value Theorem inequality?

One common misconception about the Mean Value Theorem inequality is that it applies to all functions. However, it only applies to continuous functions that are differentiable on a given interval. Another misconception is that the point where the inequality holds is always the maximum or minimum value of the function, which is not always the case.

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