Finding The Minimum Value Of F(x) = X^2 - 22x A Step-by-Step Guide
Hey guys! Today, we're diving into a cool math problem: finding the minimum value of the function . This type of problem often pops up in algebra and calculus, and it's super useful in real-world applications where we need to optimize things, like minimizing costs or maximizing profits. So, let's break it down and make sure we understand every step. We'll explore different ways to tackle this, from completing the square to using calculus, ensuring you’ve got a solid grasp of the concepts. Ready to get started?
Understanding the Function
Before we jump into solving, let's get to know our function a bit better. We're dealing with a quadratic function, specifically . Quadratic functions are those with the general form , where 'a', 'b', and 'c' are constants. In our case, , , and .
The graph of a quadratic function is a parabola, a U-shaped curve. Because the coefficient 'a' is positive (a = 1), our parabola opens upwards. This is super important because it means our function has a minimum value at the vertex of the parabola. If 'a' were negative, the parabola would open downwards, and we'd be looking for a maximum value instead. Understanding this basic shape helps us visualize what we’re trying to find.
So, what's the vertex? The vertex is the lowest point on our upward-opening parabola. It's the spot where the function transitions from decreasing to increasing. Finding the x-coordinate of the vertex is our key to solving this problem. Once we know that, we can plug it back into the function to find the minimum value, but for now, let's focus on locating that x-coordinate.
Knowing that we have a parabola opening upwards with a minimum point helps us anticipate the kind of answer we're looking for. It’s like having a roadmap before starting a journey; we know where we’re headed, which makes the trip much smoother. So, with this picture in mind, let's explore our first method for finding that minimum: completing the square.
Method 1: Completing the Square
One of the most reliable ways to find the minimum (or maximum) of a quadratic function is by completing the square. This method transforms our function into a form that makes the vertex crystal clear. The idea behind completing the square is to rewrite the quadratic expression as a perfect square plus a constant. This form directly reveals the vertex of the parabola.
Here’s how we do it step-by-step for our function :
- Focus on the and terms: We start with .
- Take half of the coefficient of the term, square it, and add it (and subtract it) inside the expression: The coefficient of our term is -22. Half of -22 is -11, and (-11)^2 is 121. So, we add and subtract 121:
- Rewrite the first three terms as a perfect square: The first three terms, , form a perfect square. They can be written as . So our function becomes:
Now, look at our transformed function: . This is in the vertex form of a quadratic equation, which is , where (h, k) is the vertex of the parabola. In our case, and .
What does this tell us? The vertex of our parabola is at the point (11, -121). Since our parabola opens upwards, the minimum value of the function occurs at this vertex. Therefore, the minimum value occurs when x = 11. And just for extra clarity, the minimum value of the function itself is -121.
Completing the square is such a powerful technique because it gives us a clear picture of the function’s behavior and directly shows us the vertex. But hey, let's not stop here! There are other ways to crack this problem, and it’s always good to have more tools in our toolbox. Let’s explore another method: using calculus.
Method 2: Using Calculus (Derivatives)
For those of you familiar with calculus, here's another fantastic way to find the minimum value of our function. Calculus provides a direct method using derivatives. The idea here is that at the minimum (or maximum) point of a function, the slope of the tangent line is zero. In other words, the derivative of the function is zero at these points.
Let’s apply this to our function, :
- Find the derivative of the function: The derivative, denoted as , tells us the slope of the function at any point. Using the power rule (where the derivative of is ), we find:
- Set the derivative equal to zero and solve for : To find the points where the slope is zero, we set :
So, we found that is a critical point where the function's slope is zero. This could be a minimum, a maximum, or a saddle point. To determine whether it’s a minimum, we can use the second derivative test.
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Find the second derivative of the function: The second derivative, , tells us about the concavity of the function. If , the function is concave up (like a smile), indicating a minimum. If , it’s concave down (like a frown), indicating a maximum. Let's find the second derivative:
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Evaluate the second derivative at our critical point: Since , which is positive, the function is concave up at . This confirms that we have a minimum at this point.
Calculus gives us a super precise way to find minimum and maximum values. By finding where the derivative is zero and using the second derivative test, we can confidently identify the nature of these critical points. Once again, we’ve found that the minimum value of the function occurs when x = 11. Isn't it cool how different methods lead us to the same answer?
Method 3: Using the Vertex Formula
Now, let’s talk about a shortcut that’s incredibly useful when dealing with quadratic functions: the vertex formula. Remember our general form for a quadratic function, ? There's a neat formula to directly calculate the x-coordinate of the vertex, which, as we know, is where our minimum (or maximum) value occurs.
The formula for the x-coordinate of the vertex (often denoted as 'h') is:
This formula comes directly from the process of completing the square, but it allows us to skip those steps and jump straight to the answer. Let's use it on our function, , where and :
Bam! Just like that, we’ve found that the x-coordinate of the vertex is 11. This method is quick and efficient, especially when you just need the x-coordinate and don't necessarily need the vertex form of the equation. It’s a great tool to have in your arsenal for exams or any situation where time is of the essence.
To find the actual minimum value of the function, we would plug x = 11 back into the original function:
So, the vertex is at (11, -121), and the minimum value of the function is -121, occurring when x is 11. The vertex formula is a fantastic shortcut, but it's always good to understand where it comes from (completing the square) and why it works. This ensures you're not just memorizing a formula but actually understanding the underlying concept.
Conclusion
Alright, guys, we've tackled the problem of finding the minimum value of using three different methods: completing the square, calculus (derivatives), and the vertex formula. Each method provides a unique perspective and reinforces the core concepts of quadratic functions. Whether you prefer the visual approach of completing the square, the precision of calculus, or the efficiency of the vertex formula, you now have multiple tools to solve similar problems.
We found that the minimum value of the function occurs when x = 11. This exercise highlights how math often provides multiple pathways to the same solution, and choosing the method that resonates best with you is key. Understanding these different approaches not only helps in solving problems but also deepens your overall understanding of mathematics.
So, next time you encounter a similar problem, remember these methods and choose the one that feels right for you. Keep practicing, and you’ll become a pro at finding minimum and maximum values in no time! Keep up the great work, and happy problem-solving!