math 1a midterm 1 (practice 1), page 6 of 6 5. (a) (15 points) using the direct definition of the derivative…

Explanation:

Step1: Recall derivative definition

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 $\sqrt{2-(x + h)}+\sqrt{2 - x}$: [ \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 $f^\prime(x)=\frac{-1}{2\sqrt{2 - x}}$, the expression under the square - root must be positive, i.e., $2 - x>0$. Solving for $x$, we get $x<2$.

Answer:

The derivative of $f(x)=\sqrt{2 - x}$ is $f^\prime(x)=\frac{-1}{2\sqrt{2 - x}}$, and the domain of $f^\prime(x)$ is $(-\infty,2)$.