Finding Critical Values For F(x) = (2/3)x^3 - 6x^2 - 80x + 10

To find the critical values of the function ${ f(x) = \frac{2}{3}x^3 - 6x^2 - 80x + 10 }$ we need to follow a systematic approach involving calculus. Critical values are crucial in understanding the behavior of a function, such as where it reaches local maxima or minima. This guide provides a detailed, step-by-step explanation to help you through the process.

1. Understanding Critical Values

Critical values, or critical points, of a function are the points in the domain where the derivative of the function is either zero or undefined. These points are essential because they indicate where the function's slope is either flat (zero derivative) or has a discontinuity (undefined derivative). At these critical points, the function may change its direction from increasing to decreasing or vice versa, making them potential locations for local maxima or minima. Essentially, finding critical values is the first step in identifying the extreme values of a function, which is a fundamental concept in calculus and optimization problems.

To truly grasp the significance of critical values, one must understand their role in the broader context of calculus. Critical values are not merely abstract mathematical points; they represent key junctures in the function's behavior. For instance, consider a roller coaster track represented by a function. The crests and troughs of the track, where the roller coaster momentarily stops changing direction, correspond to critical points. These points are where the slope of the track (the derivative of the function) is zero. Similarly, points where the track has a sharp turn or break (a discontinuity) also represent critical points, as the slope is undefined at these locations. In mathematical terms, these are the points where the function's derivative equals zero or is undefined.

The process of finding critical values involves several steps, each building upon the previous one. First, the derivative of the function must be calculated accurately. This step often requires applying various differentiation rules, such as the power rule, product rule, quotient rule, and chain rule, depending on the complexity of the function. Once the derivative is found, the next step is to identify the points where the derivative equals zero. This typically involves setting the derivative equal to zero and solving the resulting equation. The solutions to this equation are potential critical values. Additionally, one must also identify the points where the derivative is undefined. This usually occurs when the derivative involves fractions, and the denominator equals zero, or in cases of functions with discontinuities, such as piecewise functions or functions involving radicals or logarithms over intervals where the function is not defined. The combination of points where the derivative is zero and where it is undefined gives the complete set of critical values for the function.

Understanding critical values is crucial not only for mathematical analysis but also for practical applications. In optimization problems, for example, one often seeks to find the maximum or minimum value of a function subject to certain constraints. Critical points provide a roadmap for this search, as the extreme values of the function typically occur at these points or at the endpoints of the interval under consideration. In engineering, critical values can help determine the points of maximum stress on a structure or the optimal operating conditions for a machine. In economics, they can help identify points of maximum profit or minimum cost. Thus, the ability to find and interpret critical values is a powerful tool in a wide range of disciplines.

2. Calculate the Derivative

To begin, we need to find the derivative of the function f(x). The derivative, denoted as f'(x), represents the instantaneous rate of change of the function at any given point. For the given function: ${ f(x) = \frac{2}{3}x^3 - 6x^2 - 80x + 10 }$ we will apply the power rule, which states that if f(x) = ax^n, then f'(x) = nax^(n-1). This rule is fundamental to finding derivatives of polynomial functions, and its correct application is crucial for determining critical values accurately. Each term in the original function will be differentiated separately, and the results will be combined to find the overall derivative.

First, let's differentiate the term (2/3)x^3. Applying the power rule, we multiply the exponent (3) by the coefficient (2/3) and reduce the exponent by 1: ${ \frac{d}{dx} \left( \frac{2}{3}x^3 \right) = 3 \cdot \frac{2}{3}x^{3-1} = 2x^2 }$ Next, we differentiate the term -6x^2. Again, applying the power rule: ${ \frac{d}{dx} (-6x^2) = 2 \cdot (-6)x^{2-1} = -12x }$ Now, let's differentiate the term -80x. This can be thought of as -80x^1, so applying the power rule: ${ \frac{d}{dx} (-80x) = 1 \cdot (-80)x^{1-1} = -80 }$ Finally, the derivative of the constant term 10 is zero, since the derivative of any constant is zero: ${ \frac{d}{dx} (10) = 0 }$ Combining these results, we get the derivative of the function f(x): ${ f'(x) = 2x^2 - 12x - 80 }$

This derivative, f'(x) = 2x^2 - 12x - 80, is a quadratic function, and it represents the slope of the original function f(x) at any point x. The next step in finding critical values is to determine where this derivative is equal to zero or undefined. In this case, since f'(x) is a polynomial, it is defined for all real numbers, so we only need to focus on finding where it equals zero. The equation 2x^2 - 12x - 80 = 0 will give us the x-values where the slope of f(x) is zero, which are potential locations for local maxima or minima.

3. Set the Derivative to Zero

Now that we have the derivative, f'(x) = 2x^2 - 12x - 80, the next step is to find the values of x for which f'(x) = 0. These values will be potential critical values of the function f(x). Setting the derivative to zero allows us to identify the points where the function has a horizontal tangent, which often correspond to local maxima, local minima, or saddle points. Solving the quadratic equation will provide the x-coordinates of these crucial points.

We start by setting the derivative equal to zero: ${ 2x^2 - 12x - 80 = 0 }$ To simplify the equation, we can divide all terms by 2: ${ x^2 - 6x - 40 = 0 }$ This simplified quadratic equation is easier to solve. We can solve it either by factoring, completing the square, or using the quadratic formula. Factoring is often the quickest method if the quadratic expression can be factored easily. In this case, we look for two numbers that multiply to -40 and add to -6. The numbers -10 and 4 satisfy these conditions, so we can factor the quadratic equation as follows: ${ (x - 10)(x + 4) = 0 }$ Now, we set each factor equal to zero and solve for x: ${ x - 10 = 0 \quad \text{or} \quad x + 4 = 0 }$ Solving these equations gives us: ${ x = 10 \quad \text{or} \quad x = -4 }$ Thus, the values x = 10 and x = -4 are the solutions to the equation f'(x) = 0. These are the potential critical values of the function f(x). To confirm that these are indeed critical values, we need to also consider where the derivative f'(x) is undefined. However, since f'(x) = 2x^2 - 12x - 80 is a polynomial, it is defined for all real numbers. Therefore, there are no additional critical values where the derivative is undefined.

These critical values, x = 10 and x = -4, are significant because they are the x-coordinates where the tangent line to the graph of f(x) is horizontal. This means that at these points, the function may change its direction from increasing to decreasing or from decreasing to increasing. The next step would be to analyze these critical values further to determine whether they correspond to local maxima, local minima, or neither. This can be done by using the first derivative test or the second derivative test, which will provide more insight into the behavior of the function around these points.

4. Identify Critical Values

From the previous step, we found that the derivative of the function, f'(x) = 2x^2 - 12x - 80, equals zero when x = 10 and x = -4. These are the potential critical values. A critical value is a point in the domain of the function where the derivative is either zero or undefined. Since our derivative is a polynomial, it is defined for all real numbers, so we only need to consider where the derivative equals zero.

The critical values x = 10 and x = -4 are the x-coordinates where the tangent line to the graph of f(x) is horizontal. This means that at these points, the function may have a local maximum, a local minimum, or a saddle point. To determine which of these possibilities occurs at each critical value, we can use the first derivative test or the second derivative test.

The first derivative test involves examining the sign of the derivative on either side of each critical value. If the derivative changes from positive to negative at a critical value, then the function has a local maximum at that point. If the derivative changes from negative to positive, then the function has a local minimum. If the derivative does not change sign, then the point is neither a local maximum nor a local minimum (it could be a saddle point).

Alternatively, the second derivative test involves evaluating the second derivative of the function at each critical value. The second derivative, f''(x), tells us about the concavity of the function. If f''(x) > 0 at a critical value, then the function is concave up, and the critical value corresponds to a local minimum. If f''(x) < 0, then the function is concave down, and the critical value corresponds to a local maximum. If f''(x) = 0, the test is inconclusive, and we need to use another method, such as the first derivative test.

In this case, the two critical values we have identified, x = 10 and x = -4, are crucial points for understanding the behavior of the function f(x) = (2/3)x^3 - 6x^2 - 80x + 10. They divide the x-axis into three intervals: (-∞, -4), (-4, 10), and (10, ∞). By analyzing the sign of the derivative in these intervals, we can determine where the function is increasing or decreasing. Similarly, by evaluating the second derivative at these points, we can determine the concavity of the function and identify whether these points correspond to local maxima or minima.

To summarize, the critical values for the function f(x) = (2/3)x^3 - 6x^2 - 80x + 10 are x = -4 and x = 10. These points are essential for sketching the graph of the function and for solving optimization problems involving this function. The next step would be to apply either the first derivative test or the second derivative test to determine the nature of these critical points.

5. Determine Smaller Value

We have identified the critical values as x = -4 and x = 10. The question asks for the smaller value among these two. Comparing the two values, -4 is less than 10. Therefore, the smaller critical value is -4.

This value, x = -4, represents a point on the graph of the function f(x) = (2/3)x^3 - 6x^2 - 80x + 10 where the slope of the tangent line is zero. This could be a local maximum, a local minimum, or a saddle point. To determine which of these it is, we would need to perform further analysis, such as using the first or second derivative test.

The fact that x = -4 is the smaller of the two critical values also provides some insight into the function's behavior. If we were to graph the function, we would expect to see a significant change in the function's direction or concavity at x = -4. For example, if the function is increasing to the left of x = -4, it might reach a local maximum at x = -4 and then start decreasing. Alternatively, if the function is decreasing to the left of x = -4, it might reach a local minimum or a saddle point at x = -4.

In the context of optimization problems, critical values like x = -4 are crucial. When seeking to maximize or minimize a function, one often looks at the critical values and the endpoints of the interval under consideration. The maximum or minimum value of the function will occur at one of these points. Therefore, identifying the critical values is a key step in solving optimization problems.

In summary, for the function f(x) = (2/3)x^3 - 6x^2 - 80x + 10, the smaller critical value is x = -4. This value, along with the other critical value x = 10, provides valuable information about the function's behavior and potential extreme values. Further analysis using calculus techniques would be necessary to fully characterize these points and understand the function's overall shape and behavior.

By following these steps, you can effectively determine the critical values of a given function. These values are essential for understanding the behavior of the function and finding its local maxima and minima.