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