# Mean Value Theorem inequality

1. Dec 26, 2012

### mtayab1994

1. The problem statement, all variables and given/known data

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

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

3. 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.

Last edited: Dec 26, 2012
2. Dec 26, 2012

### MarneMath

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

3. Dec 26, 2012

### mtayab1994

Sorry my mistake i edited it.

4. Dec 26, 2012

### 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).

Last edited: Dec 26, 2012