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Suppose that [tex]\( f \)[/tex] is a function given as [tex]\( f(x)=\sqrt{3x+1} \)[/tex].

a. Write and simplify the difference quotient, [tex]\(\frac{f(x+h)-f(x)}{h}\)[/tex]. Note that you may need to "rationalize the numerator" as part of the simplification process.

[tex]\[
\frac{f(x+h)-f(x)}{h}=
\][/tex]
[tex]\(\square\)[/tex]

b. The derivative of the function, [tex]\( f'(x) \)[/tex], is the limit of the difference quotient as [tex]\( h \)[/tex] approaches zero.

[tex]\[
f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=
\][/tex]
[tex]\(\square\)[/tex]

Sagot :

GinnyAnsweridn
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]
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