math 1a\nmidterm 1 (practice 1), page 6 of 6\n5. (a) (15 points) using the direct definition of the…

Explanation:

Step1: Recall the direct - definition of the derivative

The direct - definition of the derivative is $f^{\prime}(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=\sqrt{2 - x}$, then $f(x + h)=\sqrt{2-(x + h)}$. So, $f^{\prime}(x)=\lim_{h\rightarrow0}\frac{\sqrt{2-(x + h)}-\sqrt{2 - x}}{h}$.

Step2: Rationalize the numerator

Multiply the numerator and denominator by the conjugate of the numerator $\sqrt{2-(x + h)}+\sqrt{2 - x}$. [ \begin{align*} f^{\prime}(x)&=\lim_{h\rightarrow0}\frac{(\sqrt{2-(x + h)}-\sqrt{2 - x})(\sqrt{2-(x + h)}+\sqrt{2 - x})}{h(\sqrt{2-(x + h)}+\sqrt{2 - x})}\ &=\lim_{h\rightarrow0}\frac{(2-(x + h))-(2 - x)}{h(\sqrt{2-(x + h)}+\sqrt{2 - x})}\ &=\lim_{h\rightarrow0}\frac{2 - x - h-2 + x}{h(\sqrt{2-(x + h)}+\sqrt{2 - x})}\ &=\lim_{h\rightarrow0}\frac{-h}{h(\sqrt{2-(x + h)}+\sqrt{2 - x})}\ &=\lim_{h\rightarrow0}\frac{-1}{\sqrt{2-(x + h)}+\sqrt{2 - x}} \end{align*} ]

Step3: Evaluate the limit

As $h\rightarrow0$, we have $f^{\prime}(x)=\frac{-1}{2\sqrt{2 - x}}$.

Step4: Find the domain of $f^{\prime}(x)$

For the function $y = f^{\prime}(x)=\frac{-1}{2\sqrt{2 - x}}$, the expression under the square - root must be positive, i.e., $2 - x>0$. Solving the inequality $2 - x>0$ gives $x<2$.

Answer:

The domain of $f^{\prime}(x)$ is $(-\infty,2)$.