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Posted on April 22, 2022 (Updated on August 4, 2025)

What is the absolute extrema in calculus?

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Absolute Extrema in Calculus: Finding the Highest and Lowest Points (the Real Deal)

Okay, so you’re diving into calculus, and sooner or later, you’re gonna run into the idea of finding the biggest and smallest values of a function. We’re not just talking about any old high or low point, but the absolute highest and lowest – the true champions of the function’s range over a specific interval. Think of it like finding the tallest and shortest person in a room, not just who’s taller than the people standing next to them. These “absolute extrema,” or global extrema if you’re feeling fancy, are super useful for all sorts of optimization problems.

So, what are absolute extrema, exactly? Simply put, they’re the ultimate maximum and minimum values a function hits within a certain domain.

  • Absolute Maximum: This is the highest point, period. A function f has an absolute maximum at x = c if f(c) is bigger than or equal to f(x) for every single x in the function’s domain. It’s the peak of the mountain.
  • Absolute Minimum: You guessed it – the lowest point. A function f has an absolute minimum at x = c if f(c) is smaller than or equal to f(x) for all x in the domain. The bottom of the valley.

Now, here’s where things get interesting. There’s this cool theorem called the Extreme Value Theorem, or EVT for short. It’s basically a guarantee that if you have a nice, continuous function on a closed interval (think of a smooth curve between two defined endpoints), you will have both an absolute maximum and an absolute minimum.

The Extreme Value Theorem (EVT): If f is continuous on a closed, bounded interval a, b, then f is guaranteed to have an absolute maximum f(c) and an absolute minimum f(d) somewhere between a and b.

But, and this is a big but, the EVT only works under certain conditions. The interval has to be closed and bounded. Open intervals or functions with breaks in them? All bets are off. You might not find those absolute extrema.

Alright, so how do you actually find these absolute extrema? That’s where the Closed Interval Method comes in. It’s a step-by-step process that makes finding those max and min values a whole lot easier.

Here’s the breakdown:

  • Find Critical Numbers: First, you need to find the critical numbers of your function f(x) within the interval (a, b). Remember those? They’re the points where the derivative f'(x) is either zero or undefined. Think of them as potential spots where the function changes direction.
  • Evaluate at Critical Numbers: Plug each of those critical numbers back into the original function f(x) to find their corresponding y-values.
  • Evaluate at Endpoints: Now, plug the endpoints of your interval, a and b, into f(x) as well.
  • Identify Absolute Extrema: Compare all the y-values you got in steps 2 and 3. The biggest one is your absolute maximum, and the smallest one is your absolute minimum. Boom!
  • Let’s walk through a couple of examples to see this in action.

    Example 1: Let’s find the absolute extrema of f(x) = x³ – 12x + 23 on the interval -5, 3.

  • Find Critical Numbers:
    • First, the derivative: f'(x) = 3x² – 12
    • Set it to zero: 3x² – 12 = 0
    • Solve: x = ±2
    • Both -2 and 2 are within our interval, so they’re both critical numbers.
  • Evaluate at Critical Numbers:
    • f(-2) = (-2)³ – 12(-2) + 23 = 39
    • f(2) = (2)³ – 12(2) + 23 = 7
  • Evaluate at Endpoints:
    • f(-5) = (-5)³ – 12(-5) + 23 = -42
    • f(3) = (3)³ – 12(3) + 23 = 14
  • Identify Absolute Extrema:
    • The biggest value is 39 (at x = -2), so that’s our absolute maximum.
    • The smallest value is -42 (at x = -5), making that our absolute minimum.
  • Example 2: Let’s try another one: g(t) = 2t³ + 3t² – 12t – 7 on the interval -4, 2.

  • Find Critical Points:
    • Derivative time: g'(t) = 6t² + 6t – 12
    • Set to zero: 6t² + 6t – 12 = 0
    • Solve: t = -2, t = 1
    • Again, both critical points are inside our interval.
  • Evaluate at Critical Points:
    • g(-2) = 2(-2)³ + 3(-2)² – 12(-2) – 7 = 13
    • g(1) = 2(1)³ + 3(1)² – 12(1) – 7 = -14
  • Evaluate at Endpoints:
    • g(-4) = 2(-4)³ + 3(-4)² – 12(-4) – 7 = -39
    • g(2) = 2(2)³ + 3(2)² – 12(2) – 7 = -3
  • Identify Absolute Extrema:
    • The absolute maximum is 13 (at t = -2).
    • The absolute minimum is -39 (at t = -4).
  • So, why should you care about all this? Because absolute extrema pop up everywhere. Seriously.

    • Business: Figuring out how to make the most money or spend the least.
    • Engineering: Designing things to be as strong or light as possible.
    • Physics: Finding the best way to throw a ball or launch a rocket.
    • Economics: Making smart investment choices.
    • Everyday Life: Even deciding the best route to work to save time!

    Understanding absolute extrema isn’t just some abstract math concept. It’s a tool that can help you solve real problems and make better decisions, no matter what you’re working on. Pretty cool, right?

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