Question

If f(x) is a polynomial of degree three with leading coefficient 1 such that f(1)=1f(1)=1, f(2)=4f(2)=4, f(3)=9f(3)=9then prove f(x)=0f(x)=0 has a root in interval (0,1)(0,1).

Answer 1

Let f(x) = ax³ + bx² + cx + d

but A/C to question , leading coefficient is 1
e.g a = 1
now,
f(1) = 1
a(1)³ + b(1)² + c(1) + d = 1
a + b + c + d = 1
b + c + d = 0 --------(1)

f(2) = 4
a(2)³ + b(2)² + c(2) + d = 4
8a + 4b + 2c + d = 4
4b + 2c + d = -4 ----------(2)

f(3) = 9
a(3)³ + b(3)² + c(3) + d = 9
27a + 9b + 3c + d = 9
9b + 3c + d = -18 ------------(3)

now, from equations (1) and (2),
3b + c = -4 -------(4)

now, from equations (2) and (3),
5b + c = -14 --------(5)
now, from equations (4) and (5)
2b = -10 , b = -5 and c = 11

now, put b and c in equation (1)
d = -6

hence, f(x) = x³ -5x² + 11x -6

now, checking in interval (0,1)
f(0) = (0)³ - 5(0)² + 11(0) - 6 < 0
f(1) = 1³ - 5(1)² + 11(1) -6 >0

here we see ,
f(0) < 0 and f(1) > 0
it means there is at least one root of f(x) in interval (0,1) .
hence, proved //