Question

If P(x) is a polynomial of degree 3 such that P(i) =(1)/(i+1)
for all i = {1,2,3,4} Then find P(5)​​

Answer 1

Answer:

P(x) = (-1/120) (x³ - 11x² + 46x - 96)

P(5) = 2/15

Step-by-step explanation:

Given, P(i) = 1/(i + 1)

For i = 1, P(1) = 1/(1 + 1) = 1/2

For i = 2, P(2) = 1/(2 + 1) = 1/3

For i = 3, P(3) = 1/(3 + 1) = 1/4

For i = 4, P(4) = 1/(4 + 1) = 1/5

Notice that 5 or any other number doesn't come in i = {1, 2,3,4}. So we can't define it for i = 5.

Let the required 3 degree polynomial be ax^3 + bx^2 + cx + d. This is the polynomial, so it must satisfy for all i(s).

=> P(1) = a(1)³ + b(1)² + c(1) + d

=> 1/2 = a + b + c + d ...(1)

For i = 2, P(2) = 1/3

=> 1/3 = a(2)³ + b(2)² + c(2) + d

=> 1/3 = 8a + 4b + 2c + d ...(2)

For i = 3, P(3) = 1/4

=> 1/4 = (3)³a + (3)²b + c(3) + d

=> 1/4 = 27a + 9b + 3c + d ...(3)

For i = 4, P(4) = 1/5

=> 1/5 = (4)³a + (4)²b + (4)c + d

=> 1/5 = 64a + 16b + 4c + d ...(4)

On solving,

(a, b, c, d) = (-1/120, 11/120, -46/120, 96/120)

*you can solve it through any method (substitution, matrix method, cramer's rule, etc)

Thus,

P(x) = ax³ + bx² + cx + d

P(x) = -x³/120 + 11x²/120 - 46x/120 + 96/120

P(x) = (-1/120) (x³ - 11x² + 46x - 96)

Hence,

P(5) = -(1/120) (5³ - 11(5)² + 46(5) - 96)

P(5) = -1/120 (-16)

P(5) = 2/15