The problem asks to find $f(x)$ as a polynomial in $x$ using the given data points and to find $f(5)$.

Given Data:

• $x: 2, 4, 6, 8$
• $y = f(x): 5, 10, 17, 29$

We need to determine the polynomial $f(x)$ that fits these data points.

Steps to Find the Polynomial:

1. Formulate the Polynomial: Assume the polynomial is of the form $f(x) = ax^3 + bx^2 + cx + d$, since there are four points, and thus a cubic polynomial might be needed.

2. Set Up Equations: Substitute each $x, y$ pair into the polynomial to set up a system of equations.

Let's solve this step by step.

Solving for the Coefficients

Given:

• $f(2) = 5$ implies $8a + 4b + 2c + d = 5$
• $f(4) = 10$ implies $64a + 16b + 4c + d = 10$
• $f(6) = 17$ implies $216a + 36b + 6c + d = 17$
• $f(8) = 29$ implies $512a + 64b + 8c + d = 29$

Let's solve this system to find $a, b, c,$ and $d$.The coefficients of the polynomial are:

• $a = \frac{1}{16}$
• $b = -\frac{1}{2}$
• $c = \frac{15}{4}$
• $d = -1$

Therefore, the polynomial is:

$f(x) = \frac{1}{16}x^3 - \frac{1}{2}x^2 + \frac{15}{4}x - 1$

Finding $f(5)$:

Now, let's substitute $x = 5$ into the polynomial to find $f(5)$.The value of $f(5)$ is:

$f(5) = \frac{209}{16} \approx 13.06$

Would you like more details, or do you have any other questions?

Related Questions:

1. How would the polynomial change if another data point were added?
2. How can you verify that this polynomial accurately fits all the given points?
3. What is the general approach to find a polynomial of degree $n$ given $n+1$ data points?
4. How would you solve this problem if the polynomial was quadratic instead of cubic?
5. What are some real-world applications of polynomial interpolation?

Tip: Always verify your polynomial by plugging the original points back into the equation to ensure accuracy!