Answer:
C. 3/8
Step-by-step explanation:
Let [tex]f'[/tex] be defined as:
[tex]f'= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]
Where:
[tex]f(x) = 3\cdot \sqrt{x}+5[/tex] and [tex]f(x+h) =3\cdot \sqrt{x+h}+5[/tex]
The definition is now expanded:
[tex]f' = \lim_{h \to 0} \frac{3\cdot \sqrt{x+h}-3\cdot \sqrt{x}}{h}[/tex]
By rationalization:
[tex]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})}[/tex]
[tex]f' = 3\cdot \lim_{h \to 0} \frac{h}{h\cdot (\sqrt{x+h}+\sqrt{x})}[/tex]
[tex]f' = 3\cdot \lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}[/tex]
[tex]f' = 3\cdot \lim_{h \to 0} \frac{1}{(x+h)^{1/2}+x^{1/2}}[/tex]
If [tex]h = 0[/tex], then:
[tex]f' = \frac{3}{2\cdot x^{1/2}}[/tex]
[tex]f' = \frac{3}{2\cdot \sqrt{x}}[/tex]
Let evaluate [tex]f'[/tex]when [tex]x = 16[/tex]:
[tex]f'(16) = \frac{3}{2\cdot \sqrt{16}}[/tex]
[tex]f'(16) = \frac{3}{8}[/tex]
Which corresponds to option C.
