Unveiling The Missing Piece: Finding F(1)

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Unveiling the Missing Piece: Finding f(1)

Hey everyone, today we're diving into a fun little math puzzle! We've got a table that shows us the values of a function, f(x)f(x), for various xx values. The kicker? The table is missing a value, and we need to find it! Let's get started. We have a set of data points, and our mission, should we choose to accept it, is to figure out what f(1)f(1) should be. This kind of problem is pretty common, so understanding how to approach it will be super helpful. Let's break down the approach and find the solution. Let's first look at what we've got. The table itself is a little treasure map, revealing how the function behaves at different points. We're given a set of input values (xx) and their corresponding output values (f(x)f(x)). The real fun begins when we start to see patterns. The most common patterns we'll encounter are linear, quadratic, and sometimes even exponential. In these situations, we'll try to find a formula that fits the data points we have, and then we'll use that formula to find the missing value.

We'll go through the various methods and how they apply to our problem. We'll start with the simplest case: a linear relationship. If the function is linear, the change in f(x)f(x) will be constant for equal changes in xx. Another approach will be to analyze if it's a quadratic function, where the function's value changes at a rate proportional to the square of the input. We'll examine the differences between the function values. This is when we can look at the differences between consecutive f(x)f(x) values. If the first differences are constant, we have a linear function. If the second differences are constant, we have a quadratic function. Let's also consider other possibilities like exponential, but it's less likely to be one of those with just a few points. However, we'll keep an open mind to explore all avenues and ensure our solution is correct. Our main goal is to find the function's value at x=1x = 1, and we'll use a systematic approach to uncover it. We'll start with the simplest assumption, then explore more complex possibilities until we have our answer. Ready to find the missing piece?

Decoding the Table: Unveiling the Function's Behavior

Alright, let's take a closer look at our table. It's like a secret code, and we're the codebreakers! The table gives us a clear view of the function's behavior at specific points. We're given the values: f(−2)=1f(-2) = 1, f(−1)=−2f(-1) = -2, and f(0)=−3f(0) = -3. This provides us with our initial data points. Now, our goal is to unveil the function's secret formula. From the data, we can start to see a pattern emerging. It looks like the function is decreasing as xx increases. This gives us our initial clues. If we want to determine the missing value, we need to first figure out the type of function. We will focus on two major possibilities: linear and quadratic. To do this, we'll calculate the differences between consecutive f(x)f(x) values. It's the key to understanding the function. If the first differences are constant, the function is linear. We can also determine the value of the slope and use a simple linear equation to find the value of f(1)f(1).

Now, let's look at the actual values:

  • When x=−2x = -2, f(x)=1f(x) = 1
  • When x=−1x = -1, f(x)=−2f(x) = -2
  • When x=0x = 0, f(x)=−3f(x) = -3

Let's calculate the differences. From f(−2)f(-2) to f(−1)f(-1), the change is −2−1=−3-2 - 1 = -3. From f(−1)f(-1) to f(0)f(0), the change is −3−(−2)=−1-3 - (-2) = -1. These differences are not the same, which means the function is not linear. Now let's determine the differences of the differences. The difference from -3 to -1 is 2. So, we'll try a quadratic function. We could assume our function to be in the form of f(x)=ax2+bx+cf(x) = ax^2 + bx + c. We can use our points to determine the values of aa, bb, and cc. The important thing is to use the data we have and apply the math. So, let's break down this function.

Solving for f(1): A Step-by-Step Approach

Now, let's roll up our sleeves and solve for f(1)f(1)! We have our data and a better understanding of how the function behaves. Given what we have, we are going to try to fit a quadratic equation to our data. With a quadratic equation, we have three unknowns, a, b, and c. We have three data points. We can write three equations using the given data. Let's start by substituting the given points into our assumed equation.

For the point (−2,1)(-2, 1), we get 1=a(−2)2+b(−2)+c1 = a(-2)^2 + b(-2) + c, which simplifies to 1=4a−2b+c1 = 4a - 2b + c.

For the point (−1,−2)(-1, -2), we get −2=a(−1)2+b(−1)+c-2 = a(-1)^2 + b(-1) + c, which simplifies to −2=a−b+c-2 = a - b + c.

For the point (0,−3)(0, -3), we get −3=a(0)2+b(0)+c-3 = a(0)^2 + b(0) + c, which simplifies to −3=c-3 = c.

Great! We now know that c=−3c = -3. Let's substitute that into the first two equations, making them 1=4a−2b−31 = 4a - 2b - 3 and −2=a−b−3-2 = a - b - 3. This simplifies to two equations: 4=4a−2b4 = 4a - 2b and 1=a−b1 = a - b. Let's solve for aa and bb. We can rearrange the second equation to get a=b+1a = b + 1. We can substitute this into the first equation to get 4=4(b+1)−2b4 = 4(b + 1) - 2b. This simplifies to 4=4b+4−2b4 = 4b + 4 - 2b, which simplifies to 0=2b0 = 2b, so b=0b = 0. Using the value of bb, we can find aa. a=b+1a = b + 1, so a=0+1a = 0 + 1, and a=1a = 1.

So, our equation is f(x)=x2−3f(x) = x^2 - 3. Now we can find the value of f(1)f(1). Substituting x=1x = 1, we get f(1)=(1)2−3f(1) = (1)^2 - 3, which gives us f(1)=1−3f(1) = 1 - 3, and the result is f(1)=−2f(1) = -2. There we have it. We cracked the code and found the missing value.

Verification and Conclusion: Ensuring Our Solution

Before we pop the champagne, let's make sure our answer is correct. We can verify our solution by looking at the equation. We know that f(x)=x2−3f(x) = x^2 - 3, and we can check the values.

  • f(−2)=(−2)2−3=4−3=1f(-2) = (-2)^2 - 3 = 4 - 3 = 1 (Correct)
  • f(−1)=(−1)2−3=1−3=−2f(-1) = (-1)^2 - 3 = 1 - 3 = -2 (Correct)
  • f(0)=(0)2−3=0−3=−3f(0) = (0)^2 - 3 = 0 - 3 = -3 (Correct)
  • f(1)=(1)2−3=1−3=−2f(1) = (1)^2 - 3 = 1 - 3 = -2 (Correct)

All of the values match our given data, and our solution is confirmed! The value of f(1)f(1) is -2. So, what did we learn today? We learned how to approach a problem where a function's value is missing. We explored different function types, and how to fit them to given data. We also learned how to use a step-by-step approach to solve for any unknown value. Keep practicing, and you'll become a function-finding expert in no time! Keep exploring, and you'll find there's a whole world of math waiting for you. This will help you to solve all kinds of problems. With this, you can apply your skills to more complex problems in the future. See ya next time!