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For what real values of X is the cubed root > than X?
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Please help! I have no idea how to do this problem!!
If X is smaller than -1, it will be true (the number will be nearer 0). If x is < 1 but > 0, it will also be true. I don't think that there are other intervals left.
Could you please state the problem more clearly? For what values of x is the cube root of what greater than x?
Assuming you mean the cubic root of x.
\[\sqrt[3]{x}>x\]
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yes
\[x-x ^{\frac{1}{3}}<0^{}\] \[x ^{\frac{1}{3}}(x ^{\frac{2}{3}}-1)>0\]
Now make pictures of the factors and see where the product is positive.
thank you !!!
yw
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