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Chapter 2: Problem 18

Find the absolute maximum and minimum values of each function over the indicated interval, and indicate the \(x\) -values at which they occur. $$f(x)=1+6 x-3 x^{2} ;[0,4]$$

Short Answer

Expert verified
The absolute maximum value is 4 at \(x = 1\) and the absolute minimum value is -23 at \(x = 4\).

Step by step solution

01

- Find the derivative

First, find the derivative of the function to determine the critical points. The given function is \[f(x) = 1 + 6x - 3x^2\]Taking the derivative, we get\[f'(x) = 6 - 6x\]
02

- Solve for critical points

Set the derivative equal to zero and solve for \(x\):\[6 - 6x = 0\]\[x = 1\]This is the critical point within the interval \([0, 4]\).
03

- Evaluate endpoints and critical points

Evaluate the function \(f(x)\) at the critical point and at the endpoints of the interval.At \(x = 0\):\[f(0) = 1 + 6(0) - 3(0^2) = 1\]At \(x = 1\):\[f(1) = 1 + 6(1) - 3(1^2) = 1 + 6 - 3 = 4\]At \(x = 4\):\[f(4) = 1 + 6(4) - 3(4^2) = 1 + 24 - 48 = -23\]
04

- Determine the absolute maximum and minimum

Compare the values obtained to determine the absolute maximum and minimum.\[f(0) = 1\]\[f(1) = 4\]\[f(4) = -23\]The absolute maximum value is 4 at \(x = 1\) and the absolute minimum value is -23 at \(x = 4\).

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Key Concepts

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

Critical Points
Critical points are crucial in finding the absolute maximum and minimum values of a function. They occur where the derivative of a function is equal to zero or does not exist. In other words, these points are where the function's slope changes. To find the critical points of a function, follow these steps:
  • First, calculate the derivative of the function.
  • Next, set the derivative equal to zero.
  • Solve the equation to find the values of x where these critical points occur.
In our exercise, the function is \[f(x) = 1 + 6x - 3x^2\]. Taking its derivative gives us \[f'(x) = 6 - 6x\]. Setting the derivative equal to zero, we get \[6 - 6x = 0 \Rightarrow x = 1\]. Therefore, x = 1 is a critical point.
Derivative
The derivative of a function helps us find the rate at which the function changes. It is a fundamental concept in calculus and helps identify the critical points. To find the derivative, apply the power rule and other differentiation rules to the function.

For the function \[f(x) = 1 + 6x - 3x^2\],

\[f'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(6x) - \frac{d}{dx}(3x^2) = 0 + 6 - 6x\],

resulting in \[f'(x) = 6 - 6x\]. By setting this equal to zero and solving for x, we discovered that x = 1 is a critical point. Understanding how to find and manipulate derivatives is key to solving many calculus problems.
Evaluate Endpoints
To find the absolute maximum and minimum values of a function over a closed interval, it is essential to evaluate the function at its endpoints, along with its critical points. This helps ensure that any maximum or minimum value within the interval is considered.

For the interval \[0, 4\]:
  • Evaluate \[f(x)\] at x = 0: \[f(0) = 1 + 6(0) - 3(0^2) = 1\].
  • Evaluate \[f(x)\] at x = 4: \[f(4) = 1 + 6(4) - 3(4^2) = 1 + 24 - 48 = -23\].
By comparing these values, we add them to the list of candidates for the absolute maximum and minimum values.
Interval Analysis
Interval analysis involves comparing the values of the function at all critical points and endpoints within the given interval. It is the last step in determining the absolute maximum and minimum of the function.

Here, we have:
\[f(0) = 1\], \[f(1) = 4\], and \[f(4) = -23\].
By analyzing these within the interval \[0, 4\]:
  • The absolute maximum value is 4 at x = 1.
  • The absolute minimum value is -23 at x = 4.
Thus, interval analysis helps conclude which points in the interval provide the maximum and minimum values of the function.

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