## University Calculus: Early Transcendentals (3rd Edition)

$g'(z) = \frac{-1}{2\sqrt (4-z)}$ $slope= g'(3) = -\frac{1}{2}$ Tangent line: $w = 2 -\frac{1}{2}(z-3)=-\frac{1}{2}z+\frac{7}{2}$
$g(z) = 1+\sqrt (4-z)$ $g'(z) = \lim\limits_{h \to 0}\frac{1+\sqrt (4-(z+h))- 1-\sqrt (4-z)}{h}$ $g'(z) = \lim\limits_{h \to 0}\frac{-h}{h(\sqrt (4-(z+h)) + \sqrt (4-z))}$ $g'(z) = \lim\limits_{h \to 0}\frac{-1}{\sqrt (4-(z+h)) + \sqrt (4-z)}$ $g'(z) = \frac{-1}{2\sqrt (4-z)}$ Slope at z = 3: $slope= g'(3) = -\frac{1}{2}$ The equation of the tangent line is: $\frac{w- 2}{z-3} = -\frac{1}{2}$ $w = 2 -\frac{1}{2}(z-3)$