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