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Question

If f (x) is a cubic polynomial such that f (1)=1,f (2)=2,f (3)=3 and f (4)=16. Find the value of f (5).

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Solution

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 get7a+3b+c=1...(5)

Solving (1) and (3), We get26a+8b+2c=2...(6)

Solving (1) and (4), We get63a+15b+3c=15...(7)

Solving (5) and (6) 6a+b=0

Solving (5) and (7) 7a+b=2

This gives, a = 2, b = -12And 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


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