Problem 2 - Calculus Single Variable

Question

In each of Exercises \(1-6,\) a function \(f\) is given. Locate each point \(c\) for which \(f(c)\) is a local extremum for \(f\). (Calculus is not needed for these exercises.) $$ f(x)=2-4(x-6)^{2} $$

Step-by-Step Solution

Answer

The local extremum is at \(x=6\), with \(f(6)=2\).

Step 1: Identify the Type of Function

The given function is of the form \( f(x) = a - b(x-c)^2 \), which represents a downward-opening parabola. This is due to the negative coefficient of the squared term \(-b\). This means the function has a maximum point at \(x = c\).

Step 2: Locate the Point of Symmetry

Since the function is a parabola, it is symmetric about the line \(x =c\). Here, the expression \((x-6)^2\) indicates that the vertex, where the local extremum occurs, is at \(x=6\).

Step 3: Calculate the Function Value at the Extremum

Substitute \(x = 6\) into the function to find the value at the extremum:\[f(6) = 2 - 4(6-6)^2 = 2 - 4(0)^2 = 2.\]Hence, the point \((6, 2)\) is the local maximum of the function.

Key Concepts

Downward-Opening Parabola

A downward-opening parabola is one where the curve looks like an upside-down U or a frown. This type of parabola can be easily recognized when the equation of the function includes a negative sign in front of the squared term. In mathematical terms, when your function is written in the form \( f(x) = a - b(x - c)^2 \) and \( b \) is positive, the parabola opens downward. This indicates that the highest point on the graph of the function is at its vertex.
But why does this happen? The squared term \( (x-c)^2 \) grows larger as \( x \) moves away from \( c \), whether positively or negatively. Because this term is subtracted from \( a \), the value of the whole function decreases as \( x \) moves away from the vertex. Hence, the vertex represents the maximum value of the function.
Downward-opening parabolas are common in various real-life situations, such as modeling scenarios where a certain peak or maximum occurs, for example, the maximum height of a projectile or optimizing profits in business models.

Local Extremum

A local extremum refers to a point on the function where it reaches a local minimum or maximum value, compared to the immediate surrounding points. In the context of parabolas, particularly downward-opening ones, we are looking at local maximum points.

• A local maximum is the highest point in a neighborhood or interval of the function.
• For a downward-opening parabola, this point occurs at the vertex.

To find a local extremum, you focus on the peak point of the parabola. In this instance, with our function \( f(x) = 2 - 4(x-6)^2 \), the local extremum occurs at \( x = 6 \).
The value of the function at this point, which is the highest it reaches in its vicinity, is \( f(6) = 2 \). Understanding this concept is crucial, as it helps in finding optimal solutions, like identifying max profit or highest volume capacity.

Vertex of a Parabola

The vertex of a parabola is a point that reflects the maximum or minimum value of a quadratic function, depending on whether it opens upwards or downwards. With downward-opening parabolas, like the one described by \( f(x) = 2 - 4(x-6)^2 \), the vertex is the point where the function attains its largest value.
The vertex can be found by looking closely at the form of the function, specifically \( (x-c)^2 \). The value of \( c \) in \( (x-c)^2 \) represents the x-coordinate of the vertex. Here, it's \( x=6 \). By plugging \( x = 6 \) back into the function, the y-coordinate is calculated, giving the vertex as the point \( (6, 2) \).

• The vertex provides symmetry to the parabola, meaning the parabola is a mirror image on either side of a vertical line through this point.

The vertex is crucial because it shows where the highest point in a downward-opening parabola occurs, serving as an axis of symmetry and assisting in solving real-world problems that require finding optimal points, like maximum heights, profits, or stresses in physical structures.