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Is the domain of f(x)=3sqrt(x-64)+5 {xIx>=4}?
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@jim_thompson5910
You're close
x>=64
oh wait, is it \[\Large f(x) = \sqrt[3]{x-64}+5\] ?
Yes =)
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so it's real
The domain of any cube root function is the set of all real numbers. This is not the same as a square root. This is because you can take the cube root of any negative number
I see! =) so it should be all real numbers?
yes
Thank you!
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yw
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