## Calculus: Early Transcendentals 8th Edition

a) The first image is the first graph of $f(x) = \sqrt{6-x}$ b) The second image is the graph of $f'(x) = \sqrt{6-x} = \frac{-1}{2\sqrt{6-x}}$ c) $f'(x) = \sqrt{6-x} = \frac{-1}{2\sqrt{6-x}}$ The domain of $f(x)$ = $(-\infty, 6]$ The domain of $f'(x)$ = $(-\infty, 6)$
a) The first image is the graph of $f(x) = \sqrt{6-x}$. We obtain this by using the transformations: reflect $\sqrt{x}$ around $y$-axis and move right 6 units. b) $f(x) = \sqrt{6-x}$ $f'(x) = \frac{d}{dx} (6-x)^{\frac{1}{2}}$ $f'(x) = \frac{1}{2} (6-x)^{\frac{1}{2} - 1} \times (-1)$ $f'(x) = \frac{1}{2} (6-x)^{-\frac{1}{2}} \times (-1)$ $f'(x) = \frac{1}{2\sqrt{6-x}} \times (-1)$ $f'(x) = \frac{-1}{2\sqrt{6-x}}$ (See second image.) (c) $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$ $f'(x) = \lim\limits_{h \to 0} \frac{f(\sqrt{6-(x+h)}) - f(\sqrt{6-x})}{h}$ $f'(x) = \lim\limits_{h \to 0} \frac{\sqrt{6-x-h} - \sqrt{6-x}}{h}$ $f'(x) = \lim\limits_{h \to 0} \frac{\sqrt{6-x-h} - \sqrt{6-x}}{h} \times \frac{\sqrt{6-x-h} + \sqrt{6-x}}{\sqrt{6-x-h} + \sqrt{6-x}}$ $f'(x) = \lim\limits_{h \to 0} \frac{(\sqrt{6-x-h})^2 - (\sqrt{6-x})^2}{h(\sqrt{6-x-h} + \sqrt{6-x})}$ $f'(x) = \lim\limits_{h \to 0} \frac{(6-x-h) - (6-x)}{h(\sqrt{6-x-h} + \sqrt{6-x})}$ $f'(x) = \lim\limits_{h \to 0} \frac{6-x-h - 6+x}{h(\sqrt{6-x-h} + \sqrt{6-x})}$ $f'(x) = \lim\limits_{h \to 0} \frac{-h}{h(\sqrt{6-x-h} + \sqrt{6-x})}$ $f'(x) = \lim\limits_{h \to 0} \frac{-1}{\sqrt{6-x-h} + \sqrt{6-x}}$ Replace $h$ for $0$ $f'(x) = \frac{-1}{\sqrt{6-x-0} + \sqrt{6-x}}$ $f'(x) = \frac{-1}{\sqrt{6-x} + \sqrt{6-x}}$ $f'(x) = \frac{-1}{2\sqrt{6-x}}$ Domain of $f(x)$ is $(-\infty, 6]$ Domain of $f'(x)$ is $(-\infty, 6)$