Question

Let f(x) be 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 us assume :
f(x)=ax3+bx2+cx+d

Then we get,
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)

Solving (1) and (2), We get

7a+3b+c=1...(5)

Solving (1) and (3), We get

26a+8b+2c=2...(6)

Solving (1) and (4), We get

63a+15b+3c=15...(7)

Solving (5) and (6)

6a+b=0

Solving (5) and (7)

7a+b=2

This gives, a = 2, b = -12

And then, c = 23 and d = - 12

Thus,f(x)=2x3−12x2+23x−12

And So, f(5)=2∗(5)3−12∗(5)2+23(5)−12=53

The required answer is f(5) = 53