Finding The 3rd Degree Interpolating Polynomial: A Step-by-Step Guide

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Hey guys! Let's dive into a super interesting problem today: figuring out the interpolating polynomial of degree 3. This might sound intimidating, but trust me, we'll break it down and make it crystal clear. We're given a table of data points, and our mission is to find a polynomial that perfectly fits these points. It's like connecting the dots, but with a smooth, curvy line instead of straight ones. So, grab your thinking caps, and let's get started!

Understanding Interpolating Polynomials

Okay, so what exactly is an interpolating polynomial? In simple terms, it's a polynomial that passes through a given set of data points. Imagine you have a scatter plot, and you want to draw a smooth curve that goes through every single point. That curve can be represented by a polynomial. The degree of the polynomial tells you how curvy it can be. A degree 1 polynomial is just a straight line, degree 2 is a parabola, degree 3 gives us a cubic curve, and so on.

When we talk about a degree 3 polynomial, we're looking for something in the form f(x) = ax³ + bx² + cx + d, where a, b, c, and d are constants that we need to figure out. Each data point we have gives us an equation, and with enough points, we can solve for these constants. The beauty of interpolating polynomials is that they allow us to estimate values between the given data points, which is super useful in many real-world applications. Think about predicting stock prices, modeling physical phenomena, or even creating smooth animations in computer graphics – interpolating polynomials are the unsung heroes behind the scenes.

Now, you might be wondering why we're focusing on a degree 3 polynomial specifically. Well, it's all about balance. A lower degree polynomial might not be flexible enough to fit all the data points accurately, while a higher degree polynomial might wiggle too much and introduce errors. A degree 3 polynomial often strikes a good balance between accuracy and smoothness, making it a popular choice for many interpolation tasks. Plus, it's complex enough to be interesting, but not so complex that it becomes a nightmare to calculate. So, let's embrace the cubic curve and see how we can find the polynomial that fits our data like a glove!

Problem Statement and Data

Let's get down to the specifics! Our mission, should we choose to accept it (and we totally do!), is to determine the interpolating polynomial of degree 3 for the data provided in the table below:

X 0 1 2 3
f(x) 1 2 15 52

What this table tells us is that when x is 0, f(x) is 1; when x is 1, f(x) is 2; when x is 2, f(x) is 15; and finally, when x is 3, f(x) is 52. These are our data points, our anchors in the sea of polynomials. We need to find that special cubic polynomial, f(x) = ax³ + bx² + cx + d, that sails smoothly through each of these points. It's like fitting a puzzle piece perfectly into its place – satisfying and elegant when it works.

To make things even more exciting, we're given a set of options to choose from. It's like a multiple-choice question, but with a twist: we need to show our work to prove why our chosen answer is the right one. Here are the options:

(A) f(x) = x³ + x² + 1 (B) f(x) = x³ + 2x² - x - 1 (C) f(x) = x³ + x - 70 (D) f(x) = 2x³ - x + 1

Now, these look like polynomials, alright, but which one is the correct polynomial? That's the million-dollar question (or, you know, the question on our math assignment). We can't just guess and hope for the best; we need a solid strategy. We need to use our knowledge of interpolating polynomials and our problem-solving skills to navigate this mathematical maze and emerge victorious. So, let's sharpen our pencils, dust off our calculators, and get ready to tackle this challenge head-on!

Setting Up the Equations

Alright, guys, let's roll up our sleeves and get to the nitty-gritty of solving this problem. The key to finding our interpolating polynomial is to use the data points we have to create a system of equations. Remember our general form for a cubic polynomial: f(x) = ax³ + bx² + cx + d. We have four unknowns (a, b, c, and d), and conveniently, we also have four data points. This is perfect because each data point will give us one equation. Think of it as a mathematical dance – each data point taking a turn to reveal a piece of the puzzle.

Let's plug in our data points one by one:

  1. When x = 0, f(x) = 1:

    • a(0)³ + b(0)² + c(0) + d = 1
    • This simplifies beautifully to d = 1. Huzzah! We've already found one of our constants. It's like finding a cheat code early in the game. This gives us a solid foundation to build upon.

  2. When x = 1, f(x) = 2:

    • a(1)³ + b(1)² + c(1) + d = 2
    • This gives us a + b + c + d = 2. Now we have an equation with four unknowns, but remember, we already know that d = 1, so we can substitute that in and simplify to a + b + c = 1.

  3. When x = 2, f(x) = 15:

    • a(2)³ + b(2)² + c(2) + d = 15
    • This translates to 8a + 4b + 2c + d = 15. Again, substituting d = 1, we get 8a + 4b + 2c = 14. We can even simplify this further by dividing everything by 2, giving us 4a + 2b + c = 7.

  4. When x = 3, f(x) = 52:

    • a(3)³ + b(3)² + c(3) + d = 52
    • This becomes 27a + 9b + 3c + d = 52. Substituting d = 1, we get 27a + 9b + 3c = 51. Dividing by 3 simplifies it to 9a + 3b + c = 17.

Now we have a system of three equations with three unknowns (a, b, and c):

  • a + b + c = 1
  • 4a + 2b + c = 7
  • 9a + 3b + c = 17

This is where the fun really begins! We've transformed our polynomial interpolation problem into a classic system of equations problem. There are several ways we can solve this system, like substitution, elimination, or even using matrices. But the goal remains the same: to find the values of a, b, and c that satisfy all three equations simultaneously. Once we have those values, we can plug them back into our polynomial equation, and voila, we'll have our interpolating polynomial. So, let's dive into the exciting world of solving systems of equations!

Solving the System of Equations

Okay, team, we've got our system of equations, and it's time to put our algebra skills to the test! We have three equations and three unknowns, which means we're in business. There are a few different ways we can tackle this, but I'm a big fan of the elimination method. It's like a mathematical game of whack-a-mole – we strategically eliminate variables until we're left with something we can solve. Let's break it down step by step:

Our equations are:

  1. a + b + c = 1
  2. 4a + 2b + c = 7
  3. 9a + 3b + c = 17

Step 1: Eliminate c from equations 1 and 2

To do this, we can subtract equation 1 from equation 2:

  • (4a + 2b + c) - (a + b + c) = 7 - 1
  • This simplifies to 3a + b = 6. Let's call this equation 4.

Step 2: Eliminate c from equations 1 and 3

Similarly, subtract equation 1 from equation 3:

  • (9a + 3b + c) - (a + b + c) = 17 - 1
  • This simplifies to 8a + 2b = 16. We can divide the whole equation by 2 to make it simpler: 4a + b = 8. Let's call this equation 5.

Now we have a new system of equations with just two variables, a and b:

  1. 3a + b = 6
  2. 4a + b = 8

Step 3: Eliminate b from equations 4 and 5

Subtract equation 4 from equation 5:

  • (4a + b) - (3a + b) = 8 - 6
  • This gives us a = 2! Awesome! We've found our first constant. It's like striking gold in a mathematical mine.

Step 4: Solve for b

Now that we know a = 2, we can plug it into either equation 4 or 5 to solve for b. Let's use equation 4:

  • 3(2) + b = 6
  • 6 + b = 6
  • So, b = 0. Another constant down! We're on a roll.

Step 5: Solve for c

We can plug the values of a and b into any of our original equations to solve for c. Let's use equation 1:

  • 2 + 0 + c = 1
  • c = -1. Fantastic! We've found all our constants.

So, after all that algebraic maneuvering, we've determined that a = 2, b = 0, c = -1, and remember from our earlier step, d = 1. Now we have all the pieces of the puzzle, and it's time to put them together to reveal our interpolating polynomial.

Constructing the Polynomial and Final Answer

Alright, guys, the moment we've been waiting for! We've done the hard work of solving for our constants, and now it's time to assemble the final product: our degree 3 interpolating polynomial. Remember the general form: f(x) = ax³ + bx² + cx + d. We've found that a = 2, b = 0, c = -1, and d = 1. So, let's plug those values in and see what we get.

f(x) = 2x³ + 0x² - 1x + 1

We can simplify this a bit by dropping the 0x² term and writing -1x as just -x:

f(x) = 2x³ - x + 1

And there you have it! This is our interpolating polynomial of degree 3 that passes through all the data points in our table. It's like building a bridge that connects all the dots, a smooth and elegant curve that captures the essence of our data.

Now, let's take a look back at the options we were given and see which one matches our result:

(A) f(x) = x³ + x² + 1 (B) f(x) = x³ + 2x² - x - 1 (C) f(x) = x³ + x - 70 (D) f(x) = 2x³ - x + 1

Drumroll, please… Option (D) is a perfect match! f(x) = 2x³ - x + 1 is indeed our interpolating polynomial. We did it! We cracked the code and found the polynomial that fits our data like a glove.

So, the final answer is:

*(D) f(x) = 2x³ - x + 1

We've successfully navigated the world of interpolating polynomials, solved a system of equations, and emerged victorious with the correct answer. Give yourselves a pat on the back, guys! You've earned it. Now, go forth and interpolate with confidence!

Repair Input Keyword

Determine the 3rd degree interpolating polynomial for the following data table:

X | 0 1 2 3 f(x) | 1 2 15 52

(A) -40 f(x) = x³ + x² + 1 (B) -40 f(x) = x³ + 2x² - x - 1 (C) 40 f(x) = x³ + x - 70 (D) -0 f(x) = 2x³ - x + 1

Title

Finding the 3rd Degree Interpolating Polynomial A Step-by-Step Guide