Answered

Connect with a global community of experts on IDNLearn.com. Our platform offers reliable and comprehensive answers to help you make informed decisions quickly and easily.

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]
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Find clear answers at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.