Problem 35 - Thomas Calculus

Question

In Exercises \(35-44,\) find the extreme values of the function and where they occur. $$ y=2 x^{2}-8 x+9 $$

Step-by-Step Solution

Answer

The minimum value is 1, occurring at \(x = 2\).

Step 1: Understand the problem

We need to find the extreme values of the quadratic function \(y = 2x^2 - 8x + 9\). Extreme values occur at points where the derivative of the function is zero or undefined. For this function, being a polynomial, the derivative will be zero at extreme points.

Step 2: Find the derivative

Let's find the derivative of the function \(y = 2x^2 - 8x + 9\). The derivative \(y'\) is calculated by differentiating each term: \(y' = 4x - 8\).

Step 3: Set derivative equal to zero

To find the points where extreme values occur, set the derivative equal to zero: \(4x - 8 = 0\). This will help us find the critical points.

Step 4: Solve for x

Solve the equation \(4x - 8 = 0\) to find the value of \(x\) where the extreme value occurs. Add 8 to both sides: \(4x = 8\). Then divide by 4: \(x = 2\).

Step 5: Find the extreme value

Substitute \(x = 2\) back into the original function \(y = 2x^2 - 8x + 9\) to find the extreme value. Calculating gives \(y = 2(2)^2 - 8(2) + 9 = 2(4) - 16 + 9 = 8 - 16 + 9 = 1\).

Step 6: Determine the nature of the extreme value

Since the function \(y = 2x^2 - 8x + 9\) is a parabola that opens upwards (coefficient of \(x^2\) is positive), the point at \(x = 2\) is a minimum point. Thus, the extreme value at this point is a minimum value.

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial function of degree two. It is typically represented in the standard form as: \( y = ax^2 + bx + c \). The coefficients \( a \), \( b \), and \( c \) are constants where \( a eq 0 \). This ensures that the function is indeed quadratic, as a zero \( a \) would reduce it to a linear function.

Quadratic functions create a distinctive curve on a graph known as a parabola. Parabolas have symmetrical U-shapes that can open upwards or downwards, depending on the sign of the leading coefficient \( a \). If \( a \) is positive, the parabola opens upwards, forming a "smile." Conversely, if \( a \) is negative, the parabola opens downwards, resembling a "frown."

In this exercise, the quadratic function is \( y = 2x^2 - 8x + 9 \). The positive coefficient \( a = 2 \) indicates that this parabola opens upwards.

Derivative

In calculus, a derivative represents the rate of change of a function with respect to a variable. It provides a way to determine how the function's output varies when there is a change in one of its inputs. For functions like quadratic ones, derivatives are essential tools to identify the behavior and characteristics of the function, such as where it reaches its extreme points.

To find the derivative of the function \( y = 2x^2 - 8x + 9 \), we differentiate each term with respect to \( x \):

• The derivative of \( 2x^2 \) is \( 4x \).
• The derivative of \( -8x \) is \( -8 \).
• The derivative of \( 9 \) is \( 0 \) since it's a constant.

Combining these results, the derivative or \( y' \) becomes \( y' = 4x - 8 \). This derivative will help us find the critical points by setting it equal to zero.

Critical Points

Critical points occur where the derivative of a function is either zero or undefined. These points are significant because they can indicate potential locations for extreme values—either a maximum or minimum point.

For the quadratic function \( y = 2x^2 - 8x + 9 \), calculating the derivative yielded \( y' = 4x - 8 \). To find the critical points, we set the derivative equal to zero:

• \( 4x - 8 = 0 \)

Solving this equation, we add 8 to both sides to get \( 4x = 8 \). Dividing both sides by 4 results in \( x = 2 \). Consequently, \( x = 2 \) is the location of a critical point, where a potential extreme value might occur.

Minimum Value

In the context of quadratic functions, minimum and maximum values refer to the lowest or highest points on the parabola, known as vertex points.

Since the parabola described by \( y = 2x^2 - 8x + 9 \) opens upwards, it will have a minimum point at its vertex. We have already determined that the critical point occurs at \( x = 2 \).

To find the minimum value of the function, substitute \( x = 2 \) back into the original equation:

• Calculate \( y = 2(2)^2 - 8(2) + 9 \).
• This simplifies to \( y = 8 - 16 + 9 \), which results in \( y = 1 \).

Therefore, the minimum value of the function is 1, occurring at the critical point \( x = 2 \). This point is where the parabola reaches its lowest point.