Problem 44 - Calculus Early Transcendentals

Question

a. Locate the critical points of \(f\) b. Use the First Derivative Test to locate the local maximum and minimum values. c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist). $$f(x)=2 x^{5}-5 x^{4}-10 x^{3}+4 \text { on }[-2,4]$$

Step-by-Step Solution

Answer

Question: Determine the absolute maximum and minimum values of the function \(f(x) = 2x^5 - 5x^4 - 10x^3 + 4\) on the interval \([-2, 4]\). Answer: The absolute maximum value is \(49\) at \(x = 3\) and the absolute minimum value is \(-40\) at \(x =-2\).

Step 1: Find the first derivative and its critical points

To find the critical points, we must first find the first derivative of \(f(x)\). \(f'(x)=\frac{d}{dx}(2x^5-5x^4-10x^3+4)\). Using the power rule, we get: \(f'(x)=10x^4 - 20x^3 - 30x^2\). Now, we need to find the critical points by finding where the first derivative is equal to 0 or undefined: \(10x^4 - 20x^3 - 30x^2 = 0\). Observe that we can factor out \(10x^2\): \(10x^2(x^2 - 2x - 3) = 0\). This equation has three solutions: \(x = 0\), \(x = -1\), and \(x = 3\). These are the critical points of \(f(x)\).

Step 2: Use the First Derivative Test to identify local maxima and minima

To apply the First Derivative Test, we will analyze the sign of \(f'(x)\) around each critical point, which will help us determine whether there is a local maximum or minimum at that point. We create a number line with our critical points and test an arbitrary number from each interval: \(f'(-2)\): choose \(x=-1.5\), then \(f'(-1.5) > 0\), so \(f'(x) > 0\) in this interval. \(f'(-1)\): choose \(x=-0.5\), then \(f'(-0.5) < 0\), so \(f'(x) < 0\) in this interval. \(f'(0)\): choose \(x=0.5\), then \(f'(0.5) > 0\), so \(f'(x) > 0\) in this interval. \(f'(3)\): choose \(x=3.5\), then \(f'(3.5) < 0\), so \(f'(x) < 0\) in this interval. From the signs of the first derivative, we can determine: - At \(x = -1\), we have a local maximum because the first derivative changes from positive to negative. - At \(x = 3\), we have a local maximum because the first derivative changes from positive to negative. - At \(x = 0\), we have a local minimum because the first derivative changes from negative to positive.

Step 3: Identify absolute maximum and minimum values on the given interval

To find the absolute maximum and minimum values, we need to evaluate \(f(x)\) at the critical points and the endpoints of the interval \([-2, 4]\): \(f(-2) = -40\) \(f(-1) = 9\) \(f(0) = 4\) \(f(3) = 49\) \(f(4) = 0\) Based on these evaluations, we can identify the absolute maximum and minimum values: - The absolute maximum value is \(49\) at \(x = 3\) - The absolute minimum value is \(-40\) at \(x = -2\)

Key Concepts

First Derivative Test

The First Derivative Test is a crucial tool in calculus for determining where a function has local maxima and minima. By taking the derivative of a function and examining where it equals zero or does not exist, you can locate its critical points. Applied to our exercise, after calculating the first derivative of the polynomial function, we look for values of x that make this derivative equal to zero, since the function is continuous and differentiable everywhere on the interval [-2,4].

By analyzing the sign of the first derivative before and after each critical point, the First Derivative Test helps us to classify these critical points. If the first derivative changes from positive to negative at a critical point, the test indicates a local maximum. Conversely, if the derivative sign swings from negative to positive, you have a local minimum. In the provided exercise, the first derivative test revealed that x = -1 and x = 3 are local maxima and x = 0 is a local minimum. This demonstrates the power of the First Derivative Test for identifying the behavior of a function.

Local Maximum and Minimum

When studying functions, identifying local maxima and minima is an essential part of understanding the function's overall behavior. These points are where the function's output value peaks or dips compared to nearby points. A local maximum is the highest point on the graph of a function within a particular interval, whereas a local minimum is the lowest point within that interval.

In our example with the polynomial function, after applying the First Derivative Test, local maxima were found at x = -1 and x = 3, while x = 0 was determined to be a local minimum. It's important to remember that local extremities are relative—it's possible for a function to have multiple local maxima and minima, as distinct from absolute ones. To fully understand the landscape of a function, observing where it crests and troughs locally provides deep insight into its fluctuating nature.

Absolute Maximum and Minimum Values

Finding a function's absolute maximum and minimum values refers to determining the highest and lowest points over the entire domain of interest. Unlike local maxima and minima, which are relative to nearby points, the absolute maximum and minimum are the supreme values a function takes on an interval.

In the context of the given exercise, to find the absolute maximum and minimum, we evaluate the original function, not just its first derivative, at all critical points as well as the endpoints. Why endpoints, you may wonder? It is because the extreme values can also occur at the boundaries of the domain. In the solved exercise, after computing the function values at critical points x = -1, x = 0, and x = 3, and at the endpoints x = -2 and x = 4, the absolute maximum value was found at x = 3, and the absolute minimum value was at x = -2. Remember: absolute extremes can occur at critical points and/or endpoints, so checking both is imperative when performing this analysis.