Question

Because of their connection with secant lines, tangentlines, and instantaneous rates, limits of the form

Answer 1

Answer:

C. 3/8

Step-by-step explanation:

Let $f'$ be defined as:

$f'= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Where:

$f(x) = 3\cdot \sqrt{x}+5$ and $f(x+h) =3\cdot \sqrt{x+h}+5$

The definition is now expanded:

$f' = \lim_{h \to 0} \frac{3\cdot \sqrt{x+h}-3\cdot \sqrt{x}}{h}$

By rationalization:

$f' = 3\cdot \lim_{h \to 0} \frac{( \sqrt{x+h}-\sqrt{x})\cdot (\sqrt{x+h}+\sqrt{x})}{h\cdot (\sqrt{x+h}+\sqrt{x})}$

$f' = 3\cdot \lim_{h \to 0} \frac{h}{h\cdot (\sqrt{x+h}+\sqrt{x})}$

$f' = 3\cdot \lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}$

$f' = 3\cdot \lim_{h \to 0} \frac{1}{(x+h)^{1/2}+x^{1/2}}$

If $h = 0$, then:

$f' = \frac{3}{2\cdot x^{1/2}}$

$f' = \frac{3}{2\cdot \sqrt{x}}$

Let evaluate $f'$when $x = 16$:

$f'(16) = \frac{3}{2\cdot \sqrt{16}}$

$f'(16) = \frac{3}{8}$

Which corresponds to option C.