Problem 1 1-20. Find (without using a calc... [FREE SOLUTION]

Chapter 3: Problem 1

1-20. Find (without using a calculator) the absolute extreme values of each function on the given interval. $$ f(x)=x^{3}-6 x^{2}+9 x+8 \text { on }[-1,2] $$

Short Answer

Expert verified

The absolute maximum is 12 and the minimum is -8.

Step by step solution

Find the critical points

To find critical points, we first need to differentiate the function $ f(x) = x^3 - 6x^2 + 9x + 8 $. The derivative is $ f'(x) = 3x^2 - 12x + 9 $. We set $ f'(x) = 0 $ to find critical points: \[ 3x^2 - 12x + 9 = 0 \]. Divide the entire equation by 3 to simplify:\[ x^2 - 4x + 3 = 0 \]. Factor the quadratic equation:\[ (x-1)(x-3) = 0 \], giving the critical points $ x = 1 $ and $ x = 3 $.

Evaluate the endpoints and critical points

Evaluate the function $ f(x) $ at the endpoints and the critical points. - At $ x = -1 $, $ f(-1) = (-1)^3 - 6(-1)^2 + 9(-1) + 8 = -1 - 6 - 9 + 8 = -8 $.- At the critical point $ x = 1 $, $ f(1) = 1^3 - 6(1)^2 + 9(1) + 8 = 1 - 6 + 9 + 8 = 12 $.- At the endpoint $ x = 2 $, $ f(2) = 2^3 - 6(2)^2 + 9(2) + 8 = 8 - 24 + 18 + 8 = 10 $.

Identify the absolute extreme values

After evaluating, we have the following values:- $ f(-1) = -8 $- $ f(1) = 12 $- $ f(2) = 10 $The absolute minimum value is $-8$ and the absolute maximum value is $12$ on the interval $[-1, 2]$.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points

Critical points are a fundamental concept in calculus, representing points where the derivative of a function is zero or undefined. These points are crucial because they often indicate locations where a function's slope changes, usually corresponding to potential minimum or maximum values. To find the critical points of a function, follow these steps:

First, take the derivative of the function. This involves applying rules of differentiation to determine the function's rate of change at any given point.
Next, solve the equation where the derivative is set to zero. For example, if the derivative is given by $f'(x)$, set $f'(x) = 0$ and solve for $x$.
Critical points can also occur where the derivative is undefined, but for most polynomial functions, you will only need to solve for where it equals zero.

In the example exercise, we differentiated $f(x) = x^3 - 6x^2 + 9x + 8$ to find $f'(x) = 3x^2 - 12x + 9$. Solving $3x^2 - 12x + 9 = 0$ gives us the critical points at $x = 1$ and $x = 3$. However, always remember to check if these critical points lie within the given interval when dealing with defined intervals.

Derivative

The derivative of a function essentially gives us the slope of the tangent line to the function's curve at any given point. It鈥檚 a measure of how the function is changing at that specific point, and it forms the backbone of calculus. Here's how derivatives function in practice:

Derivatives are used to determine critical points, as they help show where the function's rate of change is zero.
Models physical concepts such as velocity, where it provides the rate of change of position.

For example, consider finding the derivative of $f(x) = x^3 - 6x^2 + 9x + 8$. Using standard differentiation rules, the derivative is $f'(x) = 3x^2 - 12x + 9$. This function tells us where slopes are zero, indicating the critical points necessary for finding extreme values in an interval.

Extreme Values

Extreme values refer to the highest and lowest points in a given range of a function. Understanding extreme values is crucial as they indicate maxima and minima, which are essential in various real-world problems such as optimization.Here's how you identify them:

Firstly, find the critical points by setting the derivative of the function to zero and solving for $x$.
Next, evaluate the original function at these critical points, as well as the endpoints of the interval. This helps determine the absolute extreme values.
Compare these values to see which is the highest (maximum) and which is the lowest (minimum).

Returning to our example, the function $f(x) = x^3 - 6x^2 + 9x + 8$ was evaluated at $x = -1$, $x = 1$, and $x = 2$. We found the values $f(-1) = -8$, $f(1) = 12$, and $f(2) = 10$. Here, $12$ is the absolute maximum, and $-8$ is the absolute minimum on the interval $[-1, 2]$. Clearly identifying these values simplifies understanding the function's behavior over the interval.

91影视

1-20. Find (without using a calculator) the absolute extreme values of each function on the given interval. $$ f(x)=x^{3}-6 x^{2}+9 x+8 \text { on }[-1,2] $$

Short Answer

Step by step solution

Find the critical points

Evaluate the endpoints and critical points

Identify the absolute extreme values

Key Concepts

Critical Points

Derivative

Extreme Values

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Geometry

Probability and Statistics

Mechanics Maths

Theoretical and Mathematical Physics

Discrete Mathematics

Study anywhere. Anytime. Across all devices.