Problem 10 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the absolute maximum value and the absolute minimum value, if any, of each function. $$ g(x)=-x^{2}+4 x+3 $$

Step-by-Step Solution

Answer

The function g(x) = \(-\)x^2 + 4x + 3 has an absolute maximum value of 7 at x = 2, while it has no absolute minimum value.

Step 1: Find the derivative of the function

To find the critical points of the function, we first need to calculate its derivative. The derivative of g(x) = -x^2 + 4x + 3 with respect to x is: g'(x) = \(-\)2x + 4

Step 2: Find the critical points

Now, we'll find the critical points by setting the derivative equal to zero and solving for x: $$0 = -2x + 4$$ Divide both sides by -2: $$x = 2$$ We have one critical point at x = 2.

Step 3: Analyze the critical points

Now, we need to analyze the critical point and determine if it's a maximum or minimum. Since the function is a quadratic with a negative leading coefficient, we know it opens downward, meaning the vertex is the maximum point. Hence, our critical point at x = 2 represents an absolute maximum value. Now, let's find the y-coordinate of the maximum point by plugging x = 2 back into the original function: g(2) = - (2)^2 + 4(2) + 3 g(2) = - 4 + 8 + 3 g(2) = 7 Thus, the absolute maximum value of the function is 7, which occurs at x = 2.

Step 4: Determine the absolute minimum value

Since we have already found the absolute maximum value, and we know the function is a downward-opening parabola, there will be no absolute minimum value, as the function tends towards negative infinity as x goes to positive or negative infinity. In conclusion, the absolute maximum value of g(x) is 7, which occurs at x = 2, and there is no absolute minimum value.

Key Concepts

Critical Points

In calculus, critical points are essential to determine where a function reaches local maximum or minimum values. These points occur where the derivative of a function is zero or undefined. For the given quadratic function, we find the derivative and set it equal to zero to locate the critical points. This is because a derivative represents the slope of a function, and a zero derivative means the function's slope is horizontal. Hence, the point becomes a focus because either a peak or a valley could reside there.
To find the critical point in the function \( g(x) = -x^2 + 4x + 3 \), we calculated the derivative and equated it to zero:

• The derivative, \( g'(x) = -2x + 4 \), becomes zero at \( x = 2 \).

Thus, the critical point for this function is \( x = 2 \). It's crucial to evaluate these critical points further to establish their nature, whether they are maxima, minima, or neither.

Derivative

The derivative is a fundamental concept in calculus that captures the rate at which a function changes at any given point. Understanding derivatives allows us to explore various dynamic behaviors of functions. For instance, calculating the first derivative of a function helps us determine both the direction and magnitude of change. For any quadratic function like \( g(x) = -x^2 + 4x + 3 \), we can compute its derivative to reveal this dynamic behavior:

• Set \( g(x) = -x^2 + 4x + 3 \)
• The resulting derivative, \( g'(x) = -2x + 4 \), tells us how \( g(x) \) changes with \( x \).

This derivative, \( g'(x) \), is a linear function that slopes down at a constant rate determined by its coefficient, leading to a horizontal slope when \( g'(x) = 0 \). This calculation of derivatives is central to finding critical points.

Quadratic Function

Quadratic functions are polynomial functions of degree two, represented in the general form \( ax^2 + bx + c \). They graph as parabolas, which can open either upward or downward, depending on the sign of the leading coefficient. The vertex of the parabola represents either a maximum or minimum point. For the function \( g(x) = -x^2 + 4x + 3 \), the leading coefficient is negative, indicating the parabola opens downward. This feature implies that it has a highest point at its vertex. Some characteristics of quadratic functions include:

• If the leading coefficient is positive, the parabola opens upward and has a minimum vertex.
• If the leading coefficient is negative, the parabola opens downward and has a maximum vertex.

For our function, the opening direction is downward, suggesting the critical point at \( x = 2 \) is the vertex representing the maximum value of the function.

Maximum and Minimum Values

In calculus, finding maximum and minimum values involves looking at both critical points and the behavior of functions at their endpoints or at infinity. For quadratic functions, the vertex helps identify the extreme value, either maximum or minimum. An absolute maximum is the single largest value a function attains over its entire domain, while absolute minimum is the smallest.For the quadratic function \( g(x) = -x^2 + 4x + 3 \):

• Since the parabola opens downward, the vertex at \( x = 2 \) offers an absolute maximum value.
• Calculating the maximum by substituting \( x = 2 \) into the function gives \( g(2) = 7 \).
• Given the parabola descends indefinitely as \( x \) approaches positive or negative infinity, the function has no absolute minimum.

This understanding allows us to accurately interpret function behaviors and anticipate changes in dynamic systems.