Let $f(x)=(x-8)^2$.1. Give The Largest Domain On Which $f$ Is One-to-one And Non-decreasing. $\square$2. Give The Range Of \$f$[/tex\]. $\square$3. Find The Inverse Of $f$
Problem 1: Largest Domain on Which $f$ is One-to-One and Non-Decreasing
To find the largest domain on which $f$ is one-to-one and non-decreasing, we need to analyze the behavior of the function $f(x) = (x-8)^2$.
The function $f(x) = (x-8)^2$ is a quadratic function with a vertex at $(8, 0)$. Since the coefficient of the $x^2$ term is positive, the parabola opens upwards.
A function is one-to-one if it passes the horizontal line test, meaning that no horizontal line intersects the graph of the function in more than one place. For a quadratic function, this means that the function is one-to-one if it is either strictly increasing or strictly decreasing.
Since the parabola opens upwards, the function $f(x) = (x-8)^2$ is strictly increasing on the interval $(-\infty, 8]$ and strictly decreasing on the interval $[8, \infty)$. Therefore, the largest domain on which $f$ is one-to-one and non-decreasing is $(-\infty, 8]$.
Problem 2: Range of $f$
To find the range of $f$, we need to determine the set of all possible output values of the function.
Since the function $f(x) = (x-8)^2$ is a quadratic function with a vertex at $(8, 0)$, the minimum value of the function is 0, which occurs at $x = 8$. As $x$ increases or decreases from 8, the value of $f(x)$ increases without bound.
Therefore, the range of $f$ is $[0, \infty)$.
Problem 3: Find the Inverse of $f$
To find the inverse of $f$, we need to solve the equation $y = (x-8)^2$ for $x$.
We can start by taking the square root of both sides of the equation:
Since the square root of a squared quantity is the absolute value of that quantity, we can rewrite the equation as:
Now, we can solve for $x$ by isolating the absolute value expression:
Since the inverse function is a function, we need to choose one of the two possible values of $x$. We can choose the positive value of $x$, which gives us:
Therefore, the inverse of $f$ is $f^{-1}(x) = 8 + \sqrt{x}$.
Conclusion
In conclusion, we have found the largest domain on which $f$ is one-to-one and non-decreasing, which is $(-\infty, 8]$, and the range of $f$, which is $[0, \infty)$. We have also found the inverse of $f$, which is $f^{-1}(x) = 8 + \sqrt{x}$.
References
- [1] "Calculus" by Michael Spivak
- [2] "Introduction to Real Analysis" by Bartle and Sherbert
- [3] "Calculus: Early Transcendentals" by James Stewart
Discussion
- What is the significance of the vertex of the parabola in the function $f(x) = (x-8)^2$?
- How does the behavior of the function $f(x) = (x-8)^2$ change as $x$ increases or decreases from 8?
- What is the relationship between the inverse of $f$ and the original function $f$?
- How can we use the inverse of $f$ to solve equations involving $f$?
Q1: What is the largest domain on which $f$ is one-to-one and non-decreasing?
A1: The largest domain on which $f$ is one-to-one and non-decreasing is $(-\infty, 8]$.
Q2: Why is the function $f(x) = (x-8)^2$ one-to-one on the interval $(-\infty, 8]$?
A2: The function $f(x) = (x-8)^2$ is one-to-one on the interval $(-\infty, 8]$ because it is strictly increasing on this interval. This means that for any two points $x_1$ and $x_2$ in the interval $(-\infty, 8]$, if $x_1 < x_2$, then $f(x_1) < f(x_2)$.
Q3: What is the range of $f$?
A3: The range of $f$ is $[0, \infty)$.
Q4: Why is the range of $f$ $[0, \infty)$?
A4: The range of $f$ is $[0, \infty)$ because the function $f(x) = (x-8)^2$ has a minimum value of 0, which occurs at $x = 8$. As $x$ increases or decreases from 8, the value of $f(x)$ increases without bound.
Q5: How do we find the inverse of $f$?
A5: To find the inverse of $f$, we need to solve the equation $y = (x-8)^2$ for $x$. We can start by taking the square root of both sides of the equation and then solving for $x$.
Q6: What is the inverse of $f$?
A6: The inverse of $f$ is $f^{-1}(x) = 8 + \sqrt{x}$.
Q7: How can we use the inverse of $f$ to solve equations involving $f$?
A7: We can use the inverse of $f$ to solve equations involving $f$ by substituting the inverse function into the equation and solving for $x$.
Q8: What is the significance of the vertex of the parabola in the function $f(x) = (x-8)^2$?
A8: The vertex of the parabola in the function $f(x) = (x-8)^2$ is the point $(8, 0)$, which is the minimum point of the parabola. This means that the function $f(x) = (x-8)^2$ has a minimum value of 0, which occurs at $x = 8$.
Q9: How does the behavior of the function $f(x) = (x-8)^2$ change as $x$ increases or decreases from 8?
A9: As $x$ increases or decreases from 8, the value of $f(x)$ increases without bound. This means that the function $f(x) = (x-8)^2$ is strictly increasing on the interval $(-\infty, 8]$ and strictly decreasing on the interval $[8, \infty)$.
Q10: What is the relationship between the inverse of $f$ and the original function $f$?
A10: The inverse of $f$ is a function that undoes the action of the original function $f$. This means that if we apply the inverse function to the output of the original function, we get back the original input.
References
- [1] "Calculus" by Michael Spivak
- [2] "Introduction to Real Analysis" by Bartle and Sherbert
- [3] "Calculus: Early Transcendentals" by James Stewart
Discussion
- What are some other examples of functions that are one-to-one and non-decreasing?
- How can we use the concept of inverse functions to solve problems in other areas of mathematics?
- What are some real-world applications of the function $f(x) = (x-8)^2$?
- How can we use the graph of the function $f(x) = (x-8)^2$ to understand its behavior?