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

[mtayab1994]

Homework Statement

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

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

The Attempt at a Solution

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

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

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.

 

[MarneMath]

Is the inequality even true? Let x = 0, you have -1 < 1 < -1

 

[mtayab1994]

Is the inequality even true? Let x = 0, you have -1 < 1 < -1

Sorry my mistake i edited it.

 

[Dick]

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).