If f(x) is polynomial function of degree 'n'. Given that f(0)=0,f(1)=1/2,f(2)=2/3--------f(n)=n/n+1.find f(n+1)

To solve the problem, we need to analyze the polynomial function \( f(x) \) of degree \( n \) given the specific values at certain points. The values provided suggest a pattern that can help us determine the function's behavior and ultimately find \( f(n+1) \).

Understanding the Given Values

We know that:

• \( f(0) = 0 \)
• \( f(1) = \frac{1}{2} \)
• \( f(2) = \frac{2}{3} \)
• ...
• \( f(n) = \frac{n}{n+1} \)

From these values, we can observe that \( f(k) = \frac{k}{k+1} \) for \( k = 0, 1, 2, \ldots, n \). This suggests a relationship that can be expressed in a polynomial form.

Constructing the Polynomial

We can define a new function \( g(x) = (x + 1)f(x) - x \). This function \( g(x) \) is also a polynomial of degree \( n + 1 \) because it is derived from multiplying \( f(x) \) by \( (x + 1) \), which increases the degree by one.

Now, let's evaluate \( g(x) \) at the points we know:

• \( g(0) = (0 + 1)f(0) - 0 = 1 \cdot 0 - 0 = 0 \)
• \( g(1) = (1 + 1)f(1) - 1 = 2 \cdot \frac{1}{2} - 1 = 0 \)
• \( g(2) = (2 + 1)f(2) - 2 = 3 \cdot \frac{2}{3} - 2 = 0 \)
• ...
• \( g(n) = (n + 1)f(n) - n = (n + 1) \cdot \frac{n}{n + 1} - n = 0 \)

Since \( g(x) \) is a polynomial of degree \( n + 1 \) and has roots at \( x = 0, 1, 2, \ldots, n \), we can express it as:

Form of the Polynomial

Thus, we can write:

\( g(x) = c \cdot x(x - 1)(x - 2) \cdots (x - n) \)

for some constant \( c \). To find \( c \), we can evaluate \( g(-1) \):

Finding the Constant

Calculating \( g(-1) \):

\( g(-1) = (-1 + 1)f(-1) - (-1) = 0 + 1 = 1 \)

On the other hand, substituting \( -1 \) into the polynomial form gives:

\( g(-1) = c \cdot (-1)(-2)(-3) \cdots (-(n + 1)) = c \cdot (-1)^{n + 1} \cdot (n + 1)! \)

Setting these equal gives:

\( 1 = c \cdot (-1)^{n + 1} \cdot (n + 1)! \)

Thus, we find:

\( c = \frac{(-1)^{n + 1}}{(n + 1)!} \)

Finding f(n + 1)

Now, we can find \( f(n + 1) \) using \( g(n + 1) \):

\( g(n + 1) = c \cdot (n + 1)(n)(n - 1) \cdots (1) = c \cdot (n + 1)! \)

Substituting \( c \) gives:

\( g(n + 1) = \frac{(-1)^{n + 1}}{(n + 1)!} \cdot (n + 1)! = (-1)^{n + 1} \)

Now, we can find \( f(n + 1) \):

\( g(n + 1) = (n + 2)f(n + 1) - (n + 1) \)

Setting this equal to our previous result:

\( (-1)^{n + 1} = (n + 2)f(n + 1) - (n + 1) \)

Solving for \( f(n + 1) \):

\( (n + 2)f(n + 1) = (-1)^{n + 1} + (n + 1) \)

Thus:

\( f(n + 1) = \frac{(-1)^{n + 1} + (n + 1)}{n + 2} \)

Final Result

In conclusion, the value of \( f(n + 1) \) is:

\( f(n + 1) = \frac{(-1)^{n + 1} + (n + 1)}{n + 2} \)

This expression gives us the desired polynomial value based on the established pattern and calculations. If you have any further questions or need clarification on any step, feel free to ask!