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Mathematics 22 Online
OpenStudy (anonymous):

Find the derivative of f(x)=sqrt(x+2), and state the domain of f(x) and f'(x). What are the domains?

OpenStudy (anonymous):

The domain is all x values greater than -2. This is because one cannot square root a negative value

OpenStudy (anonymous):

Is that for both f(x) and f'(x)?

OpenStudy (anonymous):

*greater than or equal. you can take the square root of zero. No, hold on I'll write it out quick

OpenStudy (anonymous):

\[f(x)=\sqrt{x+2}\] HAS DOMAIN \(x+2\geq 0\) or \(x\geq -2\)

OpenStudy (anonymous):

f' the domain is x is greater than -2. This time it cannot equal -2

OpenStudy (anonymous):

\[f'(x)=\frac{1}{2\sqrt{x+2}}\] has domain \(x>0\) almost the same !

OpenStudy (anonymous):

no x can equal -1 in f'

OpenStudy (anonymous):

no i made a mistake, meant \(x>-2\) so almost the same only difference is at \(x=-2\) where \(f\) is defined and \(f'\) is not

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

domain of \(f'\): \(x>-2\)

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