Solution: To find the local extrema of the function $ P(t) = t^3 - 6t^2 + 9t + 10 $, we first compute its derivative:
["Title: How to Find Local Extrema of the Function $ P(t) = t^3 - 6t^2 + 9t + 10 $: A Step-by-Step Guide", "Understanding the behavior of functions is essential in calculus, especially when analyzing real-world phenomena like growth, investment returns, or physical systems. One key concept is identifying local extrema—points where a function reaches a local maximum or minimum. In this article, we explore how to find the local extrema of the cubic function:", "$$\nP(t) = t^3 - 6t^2 + 9t + 10\n$$", "We begin by computing the first derivative, a fundamental step in locating these critical points.", "---", "### Why Find Local Extrema?", "Local extrema help us determine where a function achieves its highest or lowest values within an interval. These points are crucial in optimization problems, economics, engineering, and data analysis. For polynomial functions like $ P(t) $, calculus provides a clear and reliable method to locate these features.", "---", "### Step 1: Compute the First Derivative", "To find local extrema, we start by calculating the derivative $ P'(t) $, which represents the rate of change of $ P(t) $:", "$$\nP(t) = t^3 - 6t^2 + 9t + 10\n$$", "Differentiate term by term:", "$$\nP'(t) = 3t^2 - 12t + 9\n$$", "---", "### Step 2: Find Critical Points", "Local extrema occur at critical points, where the derivative is zero or undefined. Since $ P'(t) $ is a polynomial, it is defined everywhere, so we solve $ P'(t) = 0 $:", "$$\n3t^2 - 12t + 9 = 0\n$$", "Divide the entire equation by 3 to simplify:", "$$\nt^2 - 4t + 3 = 0\n$$", "Factor the quadratic:", "$$\n(t - 1)(t - 3) = 0\n$$", "Thus, the critical points are:", "$$\nt = 1 \quad \ ext{and} \quad t = 3\n$$", "These values are where the slope is zero—potential candidates for local maxima or minima.", "---", "### Step 3: Apply the First or Second Derivative Test", "To determine the nature of each critical point, we use the second derivative test. First, compute $ P''(t) $:", "$$\nP''(t) = \frac{d}{dt}(3t^2 - 12t + 9) = 6t - 12\n$$", "Now evaluate $ P''(t) $ at each critical point:", "- At $ t = 1 $:\n $$\n P''(1) = 6(1) - 12 = -6 < 0\n $$\n Since the second derivative is negative, $ t = 1 $ is a local maximum.", "- At $ t = 3 $:\n $$\n P''(3) = 6(3) - 12 = 6 > 0\n $$\n Since the second derivative is positive, $ t = 3 $ is a local minimum.", "---", "### Step 4: Compute Function Values for Extrema", "To fully describe the extrema, evaluate $ P(t) $ at both $ t = 1 $ and $ t = 3 $:", "- Local maximum at $ t = 1 $:\n $$\n P(1) = (1)^3 - 6(1)^2 + 9(1) + 10 = 1 - 6 + 9 + 10 = 14\n $$", "- Local minimum at $ t = 3 $:\n $$\n P(3) = (3)^3 - 6(3)^2 + 9(3) + 10 = 27 - 54 + 27 + 10 = 10\n $$", "---", "### Summary of Results", "- Local Maximum: At $ t = 1 $, $ P(1) = 14 $\n- Local Minimum: At $ t = 3 $, $ P(3) = 10 $", "These points indicate where the function reaches its highest value locally and dips lowest locally—critical for modeling and optimization.", "---", "### Conclusion", "Finding local extrema of a function like $ P(t) = t^3 - 6t^2 + 9t + 10 $ follows a clear procedure:", "1. Compute the first derivative $ P'(t) $.\n2. Solve $ P'(t) = 0 $ to find critical points.\n3. Use the second (or first) derivative test to classify each critical point.\n4. Evaluate the original function to find exact extrema values.", "This method applies broadly to polynomial and differentiable functions, making it a foundational tool in calculus and applied mathematics.", "---", "Keywords: local extrema, solution, function analysis, derivative, $ P(t) $, $ t^3 - 6t^2 + 9t + 10 $, calculus, optimization, critical points, first derivative test, second derivative test\nMeta Description: Learn how to find local maxima and minima of $ P(t) = t^3 - 6t^2 + 9t + 10 $ using derivatives. Step-by-step guide with calculations and interpretation."]
