# Domain of the Derivative

#### Jason76

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

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

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

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

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

Some hints at problem solving strategy, please.

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#### SlipEternal

MHF Helper
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.