10. Let $f(x)=\frac{1}{\sqrt{x-2}}$.(a) Find The Domain Of $f$. Write The Answer In Interval Notation.(b) Find And Simplify $\frac{f(x+h)-f(x)}{h}$.

Introduction

In mathematics, functions are used to describe the relationship between variables. The domain of a function is the set of all possible input values for which the function is defined. In this article, we will explore the domain of a given function and then find and simplify the limit of a difference quotient.

Domain of the Function

The given function is $f(x)=\frac{1}{\sqrt{x-2}}$. To find the domain of this function, we need to determine the values of $x$ for which the function is defined.

The function is defined as long as the expression inside the square root is non-negative, i.e., $x-2 \geq 0$. Solving this inequality, we get $x \geq 2$.

However, we also need to consider the fact that the denominator of the function cannot be zero. Since the denominator is $\sqrt{x-2}$, we need to ensure that $x-2 > 0$, i.e., $x > 2$.

Therefore, the domain of the function $f$ is $[2, \infty)$.

Limit of a Difference Quotient

The difference quotient is a fundamental concept in calculus, and it is used to find the derivative of a function. The difference quotient is defined as:

$$\frac{f(x+h)-f(x)}{h}$$

To find the limit of this expression, we need to substitute the values of $f(x+h)$ and $f(x)$ into the expression.

First, let's find the value of $f(x+h)$. We have:

$$f(x+h) = \frac{1}{\sqrt{(x+h)-2}}$$

Now, let's find the value of $f(x)$. We have:

$$f(x) = \frac{1}{\sqrt{x-2}}$$

Substituting these values into the difference quotient, we get:

$$\frac{f(x+h)-f(x)}{h} = \frac{\frac{1}{\sqrt{(x+h)-2}} - \frac{1}{\sqrt{x-2}}}{h}$$

To simplify this expression, we can multiply the numerator and denominator by the conjugate of the numerator:

$$\frac{f(x+h)-f(x)}{h} = \frac{\frac{1}{\sqrt{(x+h)-2}} - \frac{1}{\sqrt{x-2}}}{h} \cdot \frac{\sqrt{(x+h)-2} \sqrt{x-2}}{\sqrt{(x+h)-2} \sqrt{x-2}}$$

Simplifying this expression, we get:

$$\frac{f(x+h)-f(x)}{h} = \frac{\sqrt{x-2} - \sqrt{(x+h)-2}}{h \sqrt{(x+h)-2} \sqrt{x-2}}$$

Now, we can take the limit as $h$ approaches zero:

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{\sqrt{x-2} - \sqrt{(x+h)-2}}{h \sqrt{(x+h)-2} \sqrt{x-2}}$$

Using the fact that $\lim_{h \to 0} \frac{\sqrt{x-2} - \sqrt{(x+h)-2}}{h} = \frac{1}{2 \sqrt{x-2}}$, we get:

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \frac{1}{2 \sqrt{x-2}} \cdot \frac{1}{\sqrt{x-2}}$$

Simplifying this expression, we get:

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \frac{1}{2(x-2)}$$

Therefore, the limit of the difference quotient is $\frac{1}{2(x-2)}$.

Conclusion

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Q: What is the domain of the function $f(x)=\frac{1}{\sqrt{x-2}}$?

A: The domain of the function $f(x)=\frac{1}{\sqrt{x-2}}$ is $[2, \infty)$. This means that the function is defined for all values of $x$ greater than or equal to 2.

Q: Why is the domain of the function $[2, \infty)$?

A: The domain of the function is $[2, \infty)$ because the expression inside the square root must be non-negative, i.e., $x-2 \geq 0$. Solving this inequality, we get $x \geq 2$. Additionally, the denominator of the function cannot be zero, so we need to ensure that $x-2 > 0$, i.e., $x > 2$.

Q: What is the limit of the difference quotient $\frac{f(x+h)-f(x)}{h}$?

A: The limit of the difference quotient $\frac{f(x+h)-f(x)}{h}$ is $\frac{1}{2(x-2)}$. This result demonstrates the importance of understanding the concept of limits in calculus.

Q: How do you find the limit of a difference quotient?

A: To find the limit of a difference quotient, you need to substitute the values of $f(x+h)$ and $f(x)$ into the expression. Then, you can simplify the expression and take the limit as $h$ approaches zero.

Q: What is the difference quotient?

A: The difference quotient is a fundamental concept in calculus, and it is used to find the derivative of a function. The difference quotient is defined as:

$$\frac{f(x+h)-f(x)}{h}$$

Q: Why is the difference quotient important?

A: The difference quotient is important because it is used to find the derivative of a function. The derivative of a function is a measure of how the function changes as the input changes.

Q: Can you give an example of how to find the limit of a difference quotient?

A: Yes, let's consider the function $f(x)=\frac{1}{\sqrt{x-2}}$. To find the limit of the difference quotient, we need to substitute the values of $f(x+h)$ and $f(x)$ into the expression. Then, we can simplify the expression and take the limit as $h$ approaches zero.

Here's an example:

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{\frac{1}{\sqrt{(x+h)-2}} - \frac{1}{\sqrt{x-2}}}{h}$$

Simplifying this expression, we get:

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \frac{1}{2(x-2)}$$

Therefore, the limit of the difference quotient is $\frac{1}{2(x-2)}$.

Conclusion

In this article, we have answered some common questions about the domain and limit of a function. We have discussed the importance of understanding the domain of a function and the concept of limits in calculus. We have also provided an example of how to find the limit of a difference quotient.