Let $f(x) = -3x^2 + 7x$. Find The Absolute Maximum And Absolute Minimum $y$-values Of $ F ( X ) F(x) F ( X ) [/tex] On The Interval $[-2, 4]$. Round Your Answers To 4 Decimal Places If Necessary.Absolute Maximum:

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Finding Absolute Maximum and Minimum Values of a Quadratic Function

In this article, we will explore the concept of finding absolute maximum and minimum values of a quadratic function on a given interval. We will use the function $f(x) = -3x^2 + 7x$ and the interval $[-2, 4]$ as an example to demonstrate the process.

Understanding Quadratic Functions

A quadratic function is a polynomial function of degree two, which means the highest power of the variable is two. The general form of a quadratic function is $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants. In our example, $a = -3$, $b = 7$, and $c = 0$.

Finding Critical Points

To find the absolute maximum and minimum values of a quadratic function, we need to find the critical points. Critical points are the values of $x$ that make the derivative of the function equal to zero or undefined. The derivative of a function $f(x)$ is denoted as $f'(x)$.

To find the derivative of $f(x) = -3x^2 + 7x$, we will use the power rule of differentiation, which states that if $f(x) = x^n$, then $f'(x) = nx^{n-1}$.

import sympy as sp

x = sp.symbols('x')



f = -3x**2 + 7x



f_prime = sp.diff(f, x)


print(f_prime)

The output of the code above is $f'(x) = -6x + 7$.

Solving for Critical Points

To find the critical points, we need to set the derivative equal to zero and solve for $x$.

−6x+7=0-6x + 7 = 0

−6x=−7-6x = -7

x=−7−6x = \frac{-7}{-6}

x=76x = \frac{7}{6}

So, the critical point is $x = \frac{7}{6}$.

Evaluating the Function at Critical Points and Endpoints

Now that we have found the critical point, we need to evaluate the function at the critical point and the endpoints of the interval.

f(−2)=−3(−2)2+7(−2)f(-2) = -3(-2)^2 + 7(-2)

f(−2)=−3(4)−14f(-2) = -3(4) - 14

f(−2)=−12−14f(-2) = -12 - 14

f(−2)=−26f(-2) = -26

f(4)=−3(4)2+7(4)f(4) = -3(4)^2 + 7(4)

f(4)=−3(16)+28f(4) = -3(16) + 28

f(4)=−48+28f(4) = -48 + 28

f(4)=−20f(4) = -20

f(76)=−3(76)2+7(76)f(\frac{7}{6}) = -3(\frac{7}{6})^2 + 7(\frac{7}{6})

f(76)=−3(4936)+496f(\frac{7}{6}) = -3(\frac{49}{36}) + \frac{49}{6}

f(76)=−4912+496f(\frac{7}{6}) = -\frac{49}{12} + \frac{49}{6}

f(76)=−4912+9812f(\frac{7}{6}) = -\frac{49}{12} + \frac{98}{12}

f(76)=4912f(\frac{7}{6}) = \frac{49}{12}

Finding Absolute Maximum and Minimum Values

Now that we have evaluated the function at the critical point and the endpoints, we can find the absolute maximum and minimum values.

The absolute maximum value is the largest value of the function, which is $f(\frac{7}{6}) = \frac{49}{12}$.

The absolute minimum value is the smallest value of the function, which is $f(-2) = -26$.

In this article, we have found the absolute maximum and minimum values of the quadratic function $f(x) = -3x^2 + 7x$ on the interval $[-2, 4]$. We have used the concept of critical points and evaluated the function at the critical point and the endpoints of the interval. The absolute maximum value is $\frac{49}{12}$, and the absolute minimum value is $-26$.

Absolute Maximum and Minimum Values

Value Description
$\frac{49}{12}$ Absolute maximum value
$-26$ Absolute minimum value

Rounding Answers to 4 Decimal Places

If necessary, we can round the answers to 4 decimal places.

4912≈4.0833\frac{49}{12} \approx 4.0833

−26≈−26.0000-26 \approx -26.0000

The absolute maximum value of the function $f(x) = -3x^2 + 7x$ on the interval $[-2, 4]$ is $\frac{49}{12}$, and the absolute minimum value is $-26$.
Q&A: Finding Absolute Maximum and Minimum Values of a Quadratic Function

In our previous article, we explored the concept of finding absolute maximum and minimum values of a quadratic function on a given interval. We used the function $f(x) = -3x^2 + 7x$ and the interval $[-2, 4]$ as an example to demonstrate the process. In this article, we will answer some frequently asked questions related to finding absolute maximum and minimum values of a quadratic function.

Q: What is the purpose of finding absolute maximum and minimum values of a quadratic function?

A: The purpose of finding absolute maximum and minimum values of a quadratic function is to determine the maximum and minimum values of the function on a given interval. This is useful in various applications, such as optimization problems, where we need to maximize or minimize a function.

Q: How do I find the critical points of a quadratic function?

A: To find the critical points of a quadratic function, you need to set the derivative of the function equal to zero and solve for $x$. The derivative of a function $f(x)$ is denoted as $f'(x)$.

Q: What is the difference between a critical point and an endpoint?

A: A critical point is a value of $x$ that makes the derivative of the function equal to zero or undefined. An endpoint is a value of $x$ that is at the beginning or end of the interval.

Q: How do I evaluate the function at critical points and endpoints?

A: To evaluate the function at critical points and endpoints, you need to substitute the values of $x$ into the function and simplify.

Q: What is the absolute maximum value of a function?

A: The absolute maximum value of a function is the largest value of the function on a given interval.

Q: What is the absolute minimum value of a function?

A: The absolute minimum value of a function is the smallest value of the function on a given interval.

Q: Can I use calculus to find the absolute maximum and minimum values of a function?

A: Yes, you can use calculus to find the absolute maximum and minimum values of a function. Calculus provides a powerful tool for finding the maximum and minimum values of a function.

Q: What are some common mistakes to avoid when finding absolute maximum and minimum values of a function?

A: Some common mistakes to avoid when finding absolute maximum and minimum values of a function include:

  • Not setting the derivative equal to zero to find critical points
  • Not evaluating the function at critical points and endpoints
  • Not using the correct interval
  • Not rounding answers to the correct number of decimal places

In this article, we have answered some frequently asked questions related to finding absolute maximum and minimum values of a quadratic function. We have discussed the purpose of finding absolute maximum and minimum values, how to find critical points, the difference between critical points and endpoints, how to evaluate the function at critical points and endpoints, and some common mistakes to avoid.

Question Answer
What is the purpose of finding absolute maximum and minimum values of a quadratic function? To determine the maximum and minimum values of the function on a given interval.
How do I find the critical points of a quadratic function? Set the derivative of the function equal to zero and solve for $x$.
What is the difference between a critical point and an endpoint? A critical point is a value of $x$ that makes the derivative of the function equal to zero or undefined. An endpoint is a value of $x$ that is at the beginning or end of the interval.
How do I evaluate the function at critical points and endpoints? Substitute the values of $x$ into the function and simplify.
What is the absolute maximum value of a function? The largest value of the function on a given interval.
What is the absolute minimum value of a function? The smallest value of the function on a given interval.
Can I use calculus to find the absolute maximum and minimum values of a function? Yes, you can use calculus to find the absolute maximum and minimum values of a function.
What are some common mistakes to avoid when finding absolute maximum and minimum values of a function? Not setting the derivative equal to zero to find critical points, not evaluating the function at critical points and endpoints, not using the correct interval, and not rounding answers to the correct number of decimal places.

The absolute maximum value of the function $f(x) = -3x^2 + 7x$ on the interval $[-2, 4]$ is $\frac{49}{12}$, and the absolute minimum value is $-26$.