Verify Lagrange's Mean value theorem Where Function :f(x)=(x-1) (x-2)(x-3) is in the interval [0,4].
Answers
Answered by
2
Given, f(x)=
x
2
−x
,x∈[1,4]
Since x
2
−x is continuous on R
and
x
2
−x
is exists in [1,4]
⇒f(x) is continuous in [1,4].
Differentiating the given function w.r.t. x
f
′
(x)=
2
1
(x
2
−x)
−1/2
.(2x−1)=
2
x
2
−x
2x−1
which exists ∀x∈R.
∴f(x) is differentiable in (1,4).
Thus, both the conditions of Lagrange's mean value theorem is satisfied therefore, ∃c in (1,4).
Such that f
′
(c)=
4−1
f(4)−f(1)
2
c
2
−c
2c−1
=
3
12
3(2c−1)=2
c
2
−c
.
12
9(4c
2
−4c+1)=48(c
2
−c)
3(4c
2
−4c+1)=16(c
2
−c)
⇒12c
2
−12c+3=16c
2
−16c
⇒4c
2
−4c−3=0
⇒(2c−3)(2c+1)=0
c=
2
3
,
2
−1
Thus, ∃ (there exist) c=
2
3
∈(1,4)
Such that f
′
(
2
3
)=
4−1
f(4)−f(1)
Hence, Lagrange's mean value theorem is verified and c=
2
3
.
Similar questions