Domain of the Derivative

Oct 2012
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\(\displaystyle g(x) = \sqrt{3 - x}\)

\(\displaystyle g(x) = (3 - x)^{1/2\)

\(\displaystyle g'(x) = -\dfrac{1}{2}(3 - x)^{-1/2}\)

\(\displaystyle g'(x) = -\dfrac{1}{2}\dfrac{1}{\sqrt{3 - x}}\)

What is the domain of the derivative, in interval notation?

Some hints at problem solving strategy, please.
 
Last edited:

SlipEternal

MHF Helper
Nov 2010
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What is the domain of \(\displaystyle g(x)\). If the domain of \(\displaystyle g(x)\) is the Riemann sphere, then the domain of \(\displaystyle g'(x)\) is also the Riemann sphere. If the domain of \(\displaystyle g(x)\) is the set of complex numbers, then the domain of \(\displaystyle g'(x)\) is the set of complex numbers minus a single point. Since you are asking for interval notation, I assume the domain of \(\displaystyle g(x)\) is a subset of the real numbers. Assuming a maximal subset of the reals, the domain of \(\displaystyle g(x)\) is all values of \(\displaystyle x\) such that \(\displaystyle 3-x \ge 0\) (square roots must be positive over the reals). So, the domain of \(\displaystyle g(x)\) would be \(\displaystyle (-\infty,3]\).

The domain of \(\displaystyle g'(x)\) must be a subset of that. Additionally, \(\displaystyle g'(x)\) is not defined when you have the square root of a negative number nor when the denominator is zero. Hence, \(\displaystyle g'(x)\) is defined over the reals for \(\displaystyle (-\infty,3)\). Since this is a subset of the domain of \(\displaystyle g(x)\), this is the correct domain.