Find The Absolute Maximum And Minimum { Y $}$-values Of { F(x) = X - 3 \ln (x) $}$ On The Interval { \left[\frac{1}{3}, 8\right]$}$, As Well As The { X $}$-value(s) Where They Occur. If There Are Multiple
Finding Absolute Maximum and Minimum Values of a Function
In calculus, finding the absolute maximum and minimum values of a function is a crucial concept. It involves identifying the points on the graph of the function where the function reaches its highest and lowest values. In this article, we will explore how to find the absolute maximum and minimum values of the function { f(x) = x - 3 \ln (x) $}$ on the interval {\left[\frac{1}{3}, 8\right]$}$, as well as the { x $}$-value(s) where they occur.
The given function is { f(x) = x - 3 \ln (x) $}$. This is a composite function, where the natural logarithm function is composed with the linear function. To find the absolute maximum and minimum values of this function, we need to analyze its behavior on the given interval.
To analyze the function, we need to find its critical points, which are the points where the function changes from increasing to decreasing or vice versa. We can find the critical points by taking the derivative of the function and setting it equal to zero.
Finding the Derivative
The derivative of the function { f(x) = x - 3 \ln (x) $}$ is given by:
{ f'(x) = 1 - \frac{3}{x} $}$
Finding Critical Points
To find the critical points, we need to set the derivative equal to zero and solve for { x $}$.
{ 1 - \frac{3}{x} = 0 $}$
Solving for { x $}$, we get:
{ x = 3 $}$
This is the only critical point of the function on the given interval.
Evaluating the Function at Critical Points
To find the absolute maximum and minimum values of the function, we need to evaluate the function at the critical points and at the endpoints of the interval.
Evaluating the Function at the Critical Point
The critical point is { x = 3 $}$. Evaluating the function at this point, we get:
{ f(3) = 3 - 3 \ln (3) $}$
Evaluating the Function at the Endpoints
The endpoints of the interval are {\frac{1}{3}$] and [8\$}. Evaluating the function at these points, we get:
{ f\left(\frac{1}{3}\right) = \frac{1}{3} - 3 \ln \left(\frac{1}{3}\right) $}$
{ f(8) = 8 - 3 \ln (8) $}$
Comparing the Values
To find the absolute maximum and minimum values of the function, we need to compare the values of the function at the critical points and at the endpoints.
Absolute Maximum Value
The absolute maximum value of the function is the largest value of the function on the given interval. Comparing the values of the function at the critical points and at the endpoints, we get:
{ f(3) = 3 - 3 \ln (3) \approx 0.4055 $}$
{ f\left(\frac{1}{3}\right) = \frac{1}{3} - 3 \ln \left(\frac{1}{3}\right) \approx 1.0986 $}$
{ f(8) = 8 - 3 \ln (8) \approx 2.6391 $}$
The largest value of the function is { f\left(\frac{1}{3}\right) = \frac{1}{3} - 3 \ln \left(\frac{1}{3}\right) \approx 1.0986 $}$. Therefore, the absolute maximum value of the function is approximately ${1.0986\$}.
Absolute Minimum Value
The absolute minimum value of the function is the smallest value of the function on the given interval. Comparing the values of the function at the critical points and at the endpoints, we get:
{ f(3) = 3 - 3 \ln (3) \approx 0.4055 $}$
{ f\left(\frac{1}{3}\right) = \frac{1}{3} - 3 \ln \left(\frac{1}{3}\right) \approx 1.0986 $}$
{ f(8) = 8 - 3 \ln (8) \approx 2.6391 $}$
The smallest value of the function is { f(3) = 3 - 3 \ln (3) \approx 0.4055 $}$. Therefore, the absolute minimum value of the function is approximately ${0.4055\$}.
In this article, we have found the absolute maximum and minimum values of the function { f(x) = x - 3 \ln (x) $}$ on the interval {\left[\frac{1}{3}, 8\right]$}$, as well as the { x $}$-value(s) where they occur. We have also analyzed the behavior of the function on the given interval and found its critical points. The absolute maximum value of the function is approximately ${1.0986\$}, and the absolute minimum value of the function is approximately ${0.4055\$}.
- [1] Calculus, 3rd edition, Michael Spivak
- [2] Calculus, 2nd edition, James Stewart
- [3] Calculus, 1st edition, Michael Spivak
Q&A: Finding Absolute Maximum and Minimum Values of a Function
In our previous article, we explored how to find the absolute maximum and minimum values of the function { f(x) = x - 3 \ln (x) $}$ on the interval {\left[\frac{1}{3}, 8\right]$}$, as well as the { x $}$-value(s) where they occur. In this article, we will answer some frequently asked questions related to finding absolute maximum and minimum values of a function.
Q: What is the difference between absolute maximum and absolute minimum values?
A: The absolute maximum value of a function is the largest value of the function on a given interval, while the absolute minimum value of a function is the smallest value of the function on a given interval.
Q: How do I find the absolute maximum and minimum values of a function?
A: To find the absolute maximum and minimum values of a function, you need to follow these steps:
- Find the critical points of the function by taking the derivative and setting it equal to zero.
- Evaluate the function at the critical points and at the endpoints of the interval.
- Compare the values of the function at the critical points and at the endpoints to find the absolute maximum and minimum values.
Q: What is a critical point?
A: A critical point is a point on the graph of a function where the function changes from increasing to decreasing or vice versa. Critical points are found by taking the derivative of the function and setting it equal to zero.
Q: How do I find the derivative of a function?
A: To find the derivative of a function, you need to use the power rule, product rule, and quotient rule of differentiation. The power rule states that if { f(x) = x^n $}$, then { f'(x) = nx^{n-1} $}$. The product rule states that if { f(x) = u(x)v(x) $}$, then { f'(x) = u'(x)v(x) + u(x)v'(x) $}$. The quotient rule states that if { f(x) = \frac{u(x)}{v(x)} $}$, then { f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} $}$.
Q: What is the significance of the absolute maximum and minimum values of a function?
A: The absolute maximum and minimum values of a function are significant because they provide information about the behavior of the function on a given interval. The absolute maximum value of a function is the largest value of the function on a given interval, while the absolute minimum value of a function is the smallest value of the function on a given interval.
Q: How do I apply the concept of absolute maximum and minimum values in real-world problems?
A: The concept of absolute maximum and minimum values can be applied in various real-world problems, such as:
- Finding the maximum and minimum values of a company's revenue and profit on a given time interval.
- Determining the maximum and minimum values of a population's growth rate on a given time interval.
- Finding the maximum and minimum values of a product's demand on a given time interval.
In this article, we have answered some frequently asked questions related to finding absolute maximum and minimum values of a function. We have also provided information about the significance of the absolute maximum and minimum values of a function and how to apply the concept in real-world problems.
- [1] Calculus, 3rd edition, Michael Spivak
- [2] Calculus, 2nd edition, James Stewart
- [3] Calculus, 1st edition, Michael Spivak