Solving for Critical Points

Oct 13, 2025 by miodeya.com 42 views

Finding the Absolute Minimum of a Quadratic Function

In this article, we will explore the concept of finding the absolute minimum of a quadratic function on a given interval. We will use the function $f(x) = x^2 - 8x + 26$ and the interval $[1, 8]$ as an example. The absolute minimum of a function is the lowest value that the function attains on a given interval.

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. The graph of a quadratic function is a parabola, which is a U-shaped curve.

The Function $f(x) = x^2 - 8x + 26$

The function $f(x) = x^2 - 8x + 26$ is a quadratic function with $a = 1$ , $b = -8$ , and $c = 26$ . To find the absolute minimum of this function on the interval $[1, 8]$ , we need to find the critical points of the function.

A critical point of a function is a point where the derivative of the function is equal to zero or undefined. To find the critical points of the function $f(x) = x^2 - 8x + 26$ , we need to find the derivative of the function.

The Derivative of $f(x)$

The derivative of $f(x) = x^2 - 8x + 26$ is $f'(x) = 2x - 8$ . To find the critical points, we need to set the derivative equal to zero and solve for $x$ .

Solving for Critical Points

Setting the derivative equal to zero, we get:

2x - 8 = 0

Solving for $x$ , we get:

x = 4

Therefore, the critical point of the function $f(x) = x^2 - 8x + 26$ is $x = 4$ .

Evaluating the Function at the Critical Point

To find the absolute minimum of the function, we need to evaluate the function at the critical point and at the endpoints of the interval.

Evaluating the Function at $x = 4$

Evaluating the function at $x = 4$ , we get:

f(4) = (4)^2 - 8(4) + 26 = 16 - 32 + 26 = 10

Evaluating the Function at the Endpoints

Evaluating the function at the endpoints of the interval, we get:

f(1) = (1)^2 - 8(1) + 26 = 1 - 8 + 26 = 19

f(8) = (8)^2 - 8(8) + 26 = 64 - 64 + 26 = 26

Comparing the Values

Comparing the values of the function at the critical point and at the endpoints, we get:

f(4) = 10 < f(1) = 19 < f(8) = 26

Therefore, the absolute minimum of the function $f(x) = x^2 - 8x + 26$ on the interval $[1, 8]$ is $f(4) = 10$ .

In this article, we found the absolute minimum of the quadratic function $f(x) = x^2 - 8x + 26$ on the interval $[1, 8]$ . We used the concept of critical points and evaluated the function at the critical point and at the endpoints of the interval. The absolute minimum of the function is $f(4) = 10$ .
Frequently Asked Questions (FAQs) about Finding the Absolute Minimum of a Quadratic Function

A: The absolute minimum of a function is the lowest value that the function attains on a given interval.

A: To find the absolute minimum of a quadratic function, you need to find the critical points of the function and evaluate the function at the critical points and at the endpoints of the interval.

A: A critical point of a function is a point where the derivative of the function is equal to zero or undefined.

A: To find the critical points of a quadratic function, you need to find the derivative of the function and set it equal to zero. Then, solve for the variable.

A: The derivative of a quadratic function is a linear function. For example, if the quadratic function is $f(x) = ax^2 + bx + c$ , then the derivative is $f'(x) = 2ax + b$ .

A: To evaluate the function at the critical points and at the endpoints of the interval, you need to plug in the values of the critical points and the endpoints into the function and simplify.

A: If the function has no critical points, then the absolute minimum of the function is the value of the function at one of the endpoints of the interval.

A: If the function has multiple critical points, then you need to evaluate the function at each critical point and at the endpoints of the interval to find the absolute minimum.

A: Yes, you can use a calculator to find the absolute minimum of a quadratic function. Most graphing calculators have a built-in function to find the minimum value of a function.

A: If you are given a quadratic function in factored form, you can use the factored form to find the critical points and evaluate the function at the critical points and at the endpoints of the interval.

A: No, you cannot use the quadratic formula to find the critical points of a quadratic function. The quadratic formula is used to find the roots of a quadratic equation, not the critical points of a quadratic function.

A: If you are given a quadratic function in vertex form, you can use the vertex form to find the critical points and evaluate the function at the critical points and at the endpoints of the interval.

In this article, we answered some frequently asked questions about finding the absolute minimum of a quadratic function. We covered topics such as critical points, derivatives, and evaluating functions at critical points and endpoints. We also discussed how to use calculators and factored forms to find the absolute minimum of a quadratic function.