Let's analyze and solve the given problem step-by-step.
Given the function [tex]\( f(x) = \sqrt{3x + 1} \)[/tex]:
### Part (a)
We need to write and simplify the difference quotient for the given function.
The difference quotient is given by:
[tex]\[ \frac{f(x + h) - f(x)}{h} \][/tex]
Substituting [tex]\( f(x) = \sqrt{3x + 1} \)[/tex] and [tex]\( f(x + h) = \sqrt{3(x + h) + 1} \)[/tex] into the difference quotient, we get:
[tex]\[ \frac{\sqrt{3(x + h) + 1} - \sqrt{3x + 1}}{h} \][/tex]
To simplify this expression, we need to rationalize the numerator:
[tex]\[ \frac{\sqrt{3(x + h) + 1} - \sqrt{3x + 1}}{h} \times \frac{\sqrt{3(x + h) + 1} + \sqrt{3x + 1}}{\sqrt{3(x + h) + 1} + \sqrt{3x + 1}} \][/tex]
This multiplication gives us:
[tex]\[ \frac{\left(\sqrt{3(x + h) + 1}\right)^2 - \left(\sqrt{3x + 1}\right)^2}{h \left(\sqrt{3(x + h) + 1} + \sqrt{3x + 1}\right)} \][/tex]
Simplifying the numerators:
[tex]\[ \left(\sqrt{3(x + h) + 1}\right)^2 = 3(x + h) + 1 \][/tex]
[tex]\[ \left(\sqrt{3x + 1}\right)^2 = 3x + 1 \][/tex]
Subtracting these, we get:
[tex]\[ 3(x + h) + 1 - (3x + 1) = 3x + 3h + 1 - 3x - 1 = 3h \][/tex]
Thus, the simplified difference quotient becomes:
[tex]\[ \frac{3h}{h \left(\sqrt{3(x + h) + 1} + \sqrt{3x + 1}\right)} = \frac{3}{\sqrt{3(x + h) + 1} + \sqrt{3x + 1}} \][/tex]
So, the simplified difference quotient is:
[tex]\[ \frac{f(x+h)-f(x)}{h}=\frac{3}{\sqrt{3(x + h) + 1} + \sqrt{3x + 1}} \][/tex]
### Part (c)
The derivative [tex]\( f'(x) \)[/tex] is the limit of the difference quotient as [tex]\( h \)[/tex] approaches 0:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \][/tex]
Using the difference quotient we have simplified, we get:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{3}{\sqrt{3(x + h) + 1} + \sqrt{3x + 1}} \][/tex]
As [tex]\( h \)[/tex] approaches 0, [tex]\( \sqrt{3(x + h) + 1} \)[/tex] approaches [tex]\( \sqrt{3x + 1} \)[/tex]. Therefore, the expression becomes:
[tex]\[ f'(x) = \frac{3}{\sqrt{3x + 1} + \sqrt{3x + 1}} = \frac{3}{2 \sqrt{3x + 1}} \][/tex]
Thus, the derivative of the function is:
[tex]\[ f'(x)= \frac{3}{2\sqrt{3x + 1}} \][/tex]
