Finding Max & Min Values Of A Function: A Step-by-Step Guide

Hey there, math enthusiasts! Today, we're diving into a fascinating calculus concept: finding the absolute maximum and absolute minimum values of a function on a given interval. This is a fundamental skill that has applications across various fields, from engineering to economics. We'll be working through a specific example to illustrate the process, making it easy to understand and apply. Let's tackle the function $f(x) = \frac{x^3}{3} - 5x^2 + 21x$ on the interval $[-2, 10]$. Don't worry if it sounds intimidating; we'll break it down step by step!

Understanding the Problem: Absolute Extrema

First things first: what exactly do we mean by absolute maximum and absolute minimum? Think of it like this: imagine a rollercoaster ride. The absolute maximum is the highest point the rollercoaster reaches on the entire track, and the absolute minimum is the lowest point. In the context of a function, the absolute maximum is the largest y-value (function value) the function attains on the specified interval, and the absolute minimum is the smallest y-value. These points are also known as the absolute extrema. The key is to remember that we're looking at the entire interval, not just a small section.

To find these extreme values, we need to consider two types of points: critical points and endpoints. Critical points are where the derivative of the function is either zero or undefined. These are potential locations for maximums and minimums. Endpoints are the boundaries of our interval. The absolute maximum or minimum could occur at either of these locations as well. The function's behavior can change dramatically at these critical points, so they are really important in finding the absolute maximum and minimum. So, we must take the derivative and determine when and where the derivative of the function is zero.

Step 1: Find the Derivative

Our first step is to find the derivative of the function $f(x) = \frac{x^3}{3} - 5x^2 + 21x$. The derivative, denoted as $f'(x)$, tells us the rate of change of the function. It's essentially the slope of the tangent line at any given point. To find the derivative, we'll apply the power rule: if $f(x) = x^n$, then $f'(x) = nx^{n-1}$. Let's differentiate term by term:

• The derivative of $\frac{x^3}{3}$ is $x^2$.
• The derivative of $-5x^2$ is $-10x$.
• The derivative of $21x$ is $21$.

Therefore, the derivative of $f(x)$ is $f'(x) = x^2 - 10x + 21$. This tells us how the original function changes its value at any point.

Step 2: Find the Critical Points

Now, we need to find the critical points. These are the x-values where the derivative $f'(x)$ is equal to zero or undefined. In our case, $f'(x) = x^2 - 10x + 21$ is a polynomial, so it's defined everywhere. Thus, we only need to solve for when the derivative is equal to zero.

So, let's set $f'(x) = 0$ and solve for x: $x^2 - 10x + 21 = 0$. This is a quadratic equation, and we can solve it by factoring, completing the square, or using the quadratic formula. Factoring is usually the easiest when it works, so let's try that. We're looking for two numbers that multiply to 21 and add up to -10. Those numbers are -3 and -7. Thus, we can factor the equation as $(x - 3)(x - 7) = 0$.

This means that the solutions are $x = 3$ and $x = 7$. These are our critical points. They are the x-values where the slope of the original function is zero (or where the function might have a maximum or minimum).

Step 3: Evaluate the Function at Critical Points and Endpoints

We have our critical points ($x = 3$ and $x = 7$) and the endpoints of our interval ($-2$ and $10$). Now, we need to evaluate the original function $f(x) = \frac{x^3}{3} - 5x^2 + 21x$ at each of these x-values. This means we'll substitute each x-value into the function and calculate the corresponding y-value.

• For $x = -2$: $f(-2) = \frac{(-2)^3}{3} - 5(-2)^2 + 21(-2) = \frac{-8}{3} - 20 - 42 = \frac{-8}{3} - 62 \approx -64.67$
• For $x = 3$: $f(3) = \frac{(3)^3}{3} - 5(3)^2 + 21(3) = 9 - 45 + 63 = 27$
• For $x = 7$: $f(7) = \frac{(7)^3}{3} - 5(7)^2 + 21(7) = \frac{343}{3} - 245 + 147 = \frac{343}{3} - 98 \approx 11.67$
• For $x = 10$: $f(10) = \frac{(10)^3}{3} - 5(10)^2 + 21(10) = \frac{1000}{3} - 500 + 210 = \frac{1000}{3} - 290 \approx 41.33$

Step 4: Determine the Absolute Maximum and Minimum

Now we compare the y-values we calculated in the previous step.

• $f(-2) \approx -64.67$
• $f(3) = 27$
• $f(7) \approx 11.67$
• $f(10) \approx 41.33$

The largest y-value is approximately 41.33, which occurs at $x = 10$. Therefore, the absolute maximum of the function on the interval $[-2, 10]$ is approximately 41.33. The smallest y-value is approximately -64.67, which occurs at $x = -2$. Therefore, the absolute minimum of the function on the interval $[-2, 10]$ is approximately -64.67.

Conclusion: Summary and Key Takeaways

So, to recap, here's how we found the absolute maximum and minimum values of $f(x) = \frac{x^3}{3} - 5x^2 + 21x$ on the interval $[-2, 10]$:

1. Found the derivative: $f'(x) = x^2 - 10x + 21$.
2. Found the critical points: $x = 3$ and $x = 7$ (by setting $f'(x) = 0$ and solving).
3. Evaluated the function at critical points and endpoints: $f(-2) \approx -64.67$, $f(3) = 27$, $f(7) \approx 11.67$, and $f(10) \approx 41.33$.
4. Determined the absolute maximum and minimum: The absolute maximum is approximately 41.33 (at $x = 10$), and the absolute minimum is approximately -64.67 (at $x = -2$).

This process is applicable to many other functions. Always remember to find the derivative, locate the critical points, evaluate the function at those points and the endpoints, and then compare the results to determine the absolute maximum and minimum. This process helps you find the highest and lowest points of a function within a specified range. Keep practicing, and you'll become a pro at finding absolute extrema in no time! Remember, these concepts are fundamental to calculus and have many applications. Mastering them will give you a solid foundation for more advanced topics.

Key Takeaways:

• The absolute maximum is the highest point on the interval, and the absolute minimum is the lowest point.
• Critical points (where the derivative is zero or undefined) and endpoints are crucial to consider.
• Always evaluate the function at critical points and endpoints to compare values.

For more in-depth practice, consider looking into similar problems or using online resources to reinforce your understanding. The more you practice, the easier this process will become.

For further exploration and examples, check out resources on Khan Academy and other reputable websites.

External Link:

For more information and practice problems, check out Khan Academy's Calculus section. Khan Academy Calculus