## Calculus: Early Transcendentals 8th Edition

a. $f'(a) = -\frac{1}{2} a^{-\frac{3}{2}}$ b. $x+2y = 3$ $x + 16y = 12$
$\lim\limits_{h \to 0} \frac{\frac{1}{\sqrt {a+h}}- \frac{1}{\sqrt a}}{h} =$ $\lim\limits_{h \to 0} \frac{\frac{\sqrt a - \sqrt {a+h}}{\sqrt a(\sqrt {a+h})}}{h} =$ $\lim\limits_{h \to 0} \frac{\frac{(\sqrt a - \sqrt {a+h}) (\sqrt a + \sqrt {a+h})}{\sqrt a(\sqrt {a+h}) (\sqrt a + \sqrt {a+h})}}{h} =$ $\lim\limits_{a \to b} \frac{a -(a+h)}{h\sqrt a (\sqrt {a+h}) (\sqrt a + \sqrt {a+h})} =$ Replace $h$ for $0$; $\lim\limits_{h \to 0} \frac{a-(a+0)}{0\sqrt a (\sqrt {a+0}) (\sqrt a + \sqrt {a+0})} =$ $\frac{-1}{\sqrt a (\sqrt a) (\sqrt a + \sqrt a)} = \frac{-1}{2a \sqrt a}= \frac{-1}{2a^{\frac{3}{2}}} = \frac{-1}{2} a^{-\frac{3}{2}}$ b. Find equations of the tangent lines at the points $(1,1)$ and $(4, \frac{1}{2})$ From part $a$ we know that $f'(a) = -\frac{1}{2} (a)^{-\frac{3}{2}}$ $f(4) = -\frac{1}{2} (4)^{-\frac{3}{2}}$ $f(4) = -\frac{1}{2} * \frac{1}{8}$ $f(4) = -\frac{1}{16}$ $16(y - \frac{1}{2}) = - (x-4)$ $16y - 8 = -x + 4$ Simplify: $x + 16y = 12$ $f(a) = -\frac{1}{2} (1)^{-\frac{3}{2}}$ $f(1) = -\frac{1}{2} * (1) = -\frac{1}{2}$ $2(y-1)= -(x-1)$ $2y-2 = -x + 1$ $x+2y = 3$ c. To make the graph of the curve and the tangents we graph the curve. $y = \frac{1}{\sqrt x}$ And both tangent lines: $x+2y = 3$ and $x + 16y = 12$