Problem 49 - Thomas Calculus in SI Units

Question

Find the extreme values (absolute and local) of the function over its natural domain, and where they occur. $$y=2 x^{2}-8 x+9$$

Step-by-Step Solution

Answer

The absolute and local minimum is at (2, 1); there is no maximum.

Step 1: Find the Derivative

Take the derivative of the function \( y = 2x^2 - 8x + 9 \). The derivative is \( y' = 4x - 8 \).

Step 2: Set the Derivative to Zero

To find the critical points, set the derivative equal to zero: \( 4x - 8 = 0 \).

Step 3: Solve for x

Solve the equation \( 4x - 8 = 0 \) to find the critical points. Adding 8 to both sides, we get \( 4x = 8 \). Dividing both sides by 4, we find \( x = 2 \).

Step 4: Evaluate the Function at the Critical Point

Find the function value at \( x = 2 \): \( y = 2(2)^2 - 8 \cdot 2 + 9 = 8 - 16 + 9 = 1 \). So, the critical point is \( (2, 1) \).

Step 5: Determine Second Derivative Sign

Calculate the second derivative \( y'' = 4 \), which is always positive, indicating that the function is concave up and thus the critical point is a local minimum.

Step 6: Identify Extreme Values on Natural Domain

Since the domain of a quadratic function is all real numbers, and it opens upwards (positive leading coefficient), the only extreme value is the local minimum at \( (2, 1) \), which is also the absolute minimum.

Key Concepts

Calculus

Calculus is a branch of mathematics that studies how things change. It allows us to compute things like slopes of curves and areas under curves. The main concepts in calculus include:

• Differentiation: Used to find the rate at which a quantity changes. This is what we do when we find the derivative of a function.
• Integration: The reverse process of differentiation, used to find areas under curves.

In problems involving extreme values, like finding the highest or lowest point on a graph, calculus plays a crucial role. By taking derivatives, you can identify critical points, which are potential candidates for these extreme values.

Derivatives

Derivatives are a fundamental tool in calculus, representing the rate of change of a function with respect to a variable. For the quadratic function given, the derivative is found by applying simple rules:

• For a term like \( ax^n \), the derivative is found using \( nax^{n-1} \).
• For constants, the derivative is zero.

Using these rules, the derivative of \( 2x^2 - 8x + 9 \) becomes \( 4x - 8 \). This expression tells us how the function \( y = 2x^2 - 8x + 9 \) changes at any point \( x \). By setting the derivative equal to zero, we locate the function's critical points, which will tell us where the function changes direction, indicating a possible max or min value.

Critical Points

Critical points occur where the derivative of a function equals zero or does not exist. These points can signify potential local maxima, minima, or points of inflection:

• Solving \( y' = 0 \) helps us find where the function's slope is zero, meaning the curve changes direction.
• For the quadratic \( y = 2x^2 - 8x + 9 \), we solve \( 4x - 8 = 0 \) to find \( x = 2 \).

At this point, the slope changes from negative to positive, indicating a minimum point at \( x = 2 \). Checking the function's value at this point gives us \( y = 1 \), confirming a critical point at \( (2, 1) \). Evaluating the second derivative, \( y'' = 4 \), confirms a local minimum since it is positive, indicating the curve is concave up.

Quadratic Function

A quadratic function is a polynomial of degree 2 and can be written in the standard form \( y = ax^2 + bx + c \). Its graph is a parabola that can open upwards or downwards:

• Upwards Opening: Happens when the coefficient \( a \) is positive, like in our function \( y = 2x^2 - 8x + 9 \).
• Vertex Form: The vertex is the peak or trough of the parabola, revealing important features like the minimum or maximum value of the function.

For the given function, since \( a = 2 \) (positive), the parabola opens upwards, ensuring there is a minimum point. By completing the square or using calculus, we find this vertex at \( (2, 1) \), showing it's the lowest point on the graph. Hence, this local minimum is also the absolute minimum as the arms of the parabola stretch infinitely upward.