Finding Max & Min Values: A Calculus Guide

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Hey guys! Let's dive into a classic calculus problem: finding the absolute maximum and minimum values of a function over a specific interval. This is super useful for all sorts of real-world applications, from optimizing designs to understanding how things change. We'll break down the process step-by-step, making it easy to understand, even if you're just starting out with calculus. So, let's get started, shall we?

Understanding the Problem: Absolute Maxima and Minima

Okay, so what exactly are we trying to find? We're looking for the absolute maximum and absolute minimum values of a function, f(x)f(x), within a given interval, which in our case is [−4,4][-4, 4]. Think of it like this: imagine the graph of the function as a roller coaster. The absolute maximum is the highest point the roller coaster reaches within the interval, and the absolute minimum is the lowest point. These are the highest and lowest y-values the function takes on within that specific range of x-values.

Here's the deal: to find these, we need to consider a couple of key places:

  1. Critical Points: These are the x-values where the derivative of the function, f′(x)f'(x), is either equal to zero or undefined. At these points, the function might have a peak (maximum), a valley (minimum), or a point where the slope changes direction (inflection point). Finding critical points is like identifying potential spots where our roller coaster might have a high or low point.
  2. Endpoints of the Interval: Since we're only looking at the function within the interval [−4,4][-4, 4], we also need to check the y-values at the endpoints, x=−4x = -4 and x=4x = 4. These endpoints could be the highest or lowest points within the interval, even if the function doesn't have a critical point there.

Basically, the absolute maximum and minimum will always occur either at a critical point inside the interval or at one of the interval's endpoints. Pretty straightforward, right? We'll apply this logic to the specific function f(x)=(x+1)23−6f(x)=(x+1)^{\frac{2}{3}}-6 over the interval [−4,4][-4, 4] and find the absolute maximum and minimum values.

Now, before we jump into the math, it's worth stressing the importance of these concepts. Understanding maxima and minima is crucial in optimization problems. For instance, imagine you're designing a container and want to minimize the amount of material used. You'd use calculus to find the dimensions that achieve the minimum surface area (the absolute minimum in this scenario). Or, think about economics: businesses use these concepts to find the production level that maximizes profit. It all boils down to finding the best (maximum) or the most efficient (minimum) value of something. So, by solving this problem, we're building the foundation for tackling many other interesting and useful problems.

Let's get cracking with our function! We'll start with finding those critical points.

Step-by-Step Solution: Finding the Maximum and Minimum Values

Alright, let's get our hands dirty with the actual problem. We have the function f(x)=(x+1)23−6f(x) = (x+1)^{\frac{2}{3}} - 6 and the interval [−4,4][-4, 4]. Here’s how we'll find the absolute maximum and minimum:

  1. Find the Derivative: The first step is to find the derivative of the function, f′(x)f'(x). This tells us about the slope of the function at any given point. Using the power rule and chain rule, we get:

    f′(x)=23(x+1)−13=23x+13f'(x) = \frac{2}{3}(x+1)^{-\frac{1}{3}} = \frac{2}{3\sqrt[3]{x+1}}

    Important: Remember that the derivative gives us the slope of the original function. Where the derivative is zero or undefined, the original function might have a maximum, minimum, or a point where the graph changes direction.

  2. Find Critical Points: Now, we need to find the critical points. These are the x-values where f′(x)=0f'(x) = 0 or where f′(x)f'(x) is undefined.

    • Where f'(x) = 0: Set the derivative equal to zero and solve for x: 23x+13=0\frac{2}{3\sqrt[3]{x+1}} = 0. This equation has no solution because the numerator is a constant (2), and cannot be zero.
    • Where f'(x) is undefined: The derivative is undefined when the denominator is equal to zero: 3x+13=03\sqrt[3]{x+1} = 0. Solving for x, we get x=−1x = -1. This is our critical point! At x=−1x = -1, the function has a vertical tangent, indicating a possible maximum or minimum (or a change in the function's direction).

  3. Evaluate the Function at Critical Points and Endpoints: We'll plug the critical point and the endpoints of the interval into the original function, f(x)f(x), to find their corresponding y-values.

    • f(−4)=(−4+1)23−6=(−3)23−6≈−4.08f(-4) = (-4+1)^{\frac{2}{3}} - 6 = (-3)^{\frac{2}{3}} - 6 \approx -4.08
    • f(−1)=(−1+1)23−6=0−6=−6f(-1) = (-1+1)^{\frac{2}{3}} - 6 = 0 - 6 = -6
    • f(4)=(4+1)23−6=(5)23−6≈−2.16f(4) = (4+1)^{\frac{2}{3}} - 6 = (5)^{\frac{2}{3}} - 6 \approx -2.16

  4. Identify the Maximum and Minimum: Now, compare the y-values we calculated. The largest y-value is the absolute maximum, and the smallest y-value is the absolute minimum.

    • Absolute Maximum: Approximately -2.16, which occurs at x=4x = 4.
    • Absolute Minimum: -6, which occurs at x=−1x = -1.

And there you have it, guys! We've successfully found the absolute maximum and minimum values of the function over the given interval. The process involves finding the derivative, identifying critical points, evaluating the function, and then comparing the values to pinpoint the highest and lowest points. See? Not so bad, right?

Visualizing the Solution: Understanding the Graph

To really solidify our understanding, let's take a look at the graph of f(x)=(x+1)23−6f(x) = (x+1)^{\frac{2}{3}} - 6 over the interval [−4,4][-4, 4]. Visualizing the graph helps us see what we've calculated and why those points are the maximum and minimum values. Imagine plotting this function on a coordinate plane, with x-values ranging from -4 to 4. You'd notice the following:

  1. Shape of the Curve: The function has a shape that's somewhat like a parabola, but it's been