Question

Let f(x) be a cubic polynomial such that f(1)=1,f(2)=2,f(3)=3 and f(4)=16.find the value of f(5)

Answer 1

Let f(x)=ax³+bx²+cx+d
f(1)=a+b+c+d=1------------------(1)
f(2)=8a+4b+2c+d=2-------------(2)
f(3)=27a+9b+3c+d=3-----------(3)
f(4)=64a+16b+4c+d=16--------(4)
subtracting (1) from (2),(3) and (4) respectively we have,
7a+3b+c=1-------------------(5)
26a+8b+2c=2----------------(6)
63a+15b+3c=15-------------(7)
multiplying (5) with 2 and subtracting from (6) and multiplying (5) with 3 and subtracting from (7) we have,
12a+2b=0-------(8)
42a+6b=12-----(9)
multiplying (8) with 3 and (9) with 1 and subtracting we have,
36a+6b=0
42a+6b=12
6a=12
or, a =12
then, from (8) we have, 2b=-12a=-12×12=-144
or, b=-144/2=-72
then from (5), c=1-3b-7a=1-3×(-72)-7×12=1+216-84=217-84=133
then from (1), d=1-a-b-c=1-12-(-72)-133=1-12+72-133=73-145=-72
then, f(x)=12x³-72x²+133x-72
∴, f(5)=12×5³-72×5²+133×5-72
=12×125-72×25+665-72
=1500-1800+593
=2093-1800
=293

Answer 2

Answer:

Given:

f(1) = 1

f(2) = 2

f(3) = 3

f(4) = 16

To find : f(5)

Solution :

Assume another function g(x) such that g(x) = f(x) - x

Hence,

g(1) = 0 = x-1 ..... (1)

g(2) = 0 = x-2 ...... (2)

g(3) = 0 = x-3 ...... (3)

From (1,2,3), we can say that

g(x) = k (x-1)(x-2)(x-3) ...... (4)

f(4) = 16 .... Given

g(4) = 16 - 4 = 12

Therefore, g(4) = k(4-1)(4-2)(4-3)

12= k(4-1)(4-2)(4-3)

K=2

Equation 4 becomes

g(x) = 2 (x-1)(x-2)(x-3)

g(5) = 2(5-1)(5-2)(5-3)

g(5) = 2*4*3*2

g(5) = 48

But we know that

g(x) = f(x) - x

g(5) = f(5) - 5

f(5) = 48 + 5

Thus, f(5) = 53