Question

f(X) = (x-1)(x-2)(x-3) in {0,4} in legranger theorem​

Answer 1

f(x)=(x−1)(x−2)(x−3)=(x

3

−6x

2

+11x−6)

f(x) is a cubic polynomial function and continuous at (0,4)

f'(x) =3x

2

−12x+11 exists

So f'(x) is differentiable at (0,4)

Now f'(c)=3c

2

−12c+11

Now f'(c)=

4−0

f(4)−f(0)

3c

2

−12c+11=

4

6−(−6)

=3

Solving we get c=2

−

+

3

2

=3.154 or 0.84