- #1
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.
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