Verify the rolles theorem for f(x)=1-(x-1)^(2/3) in [0,2]
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Answer: f(x)=(x−1)(x−2)(x−3) in the interval [0,4]
We know that a polynomial function is continuous everywhere and also differentiable.
So f(x) being a polynomial is continuous and differentiable on (0,4)
So there must exist at least one real number c∈(0,4) such that
f
′
(c)=
4−0
f(4)−f(0)
=
4
f(4)−f(0)
We have f(x)=(x−1)(x−2)(x−3)
=x
3
−(1+2+3)x
2
+(2+6+3)x−6
∴f(x)=x
3
−6x
2
+11x−6
f(0)=−6
f(4)=(4)
3
−6(4)
2
+11(4)−6=64−96+44−6=6
f
′
(x)=3x
2
−12x+11
f
′
(c)=3c
2
−12c+11
∴3c
2
−12c+11=
4−0
6−(−6)
⇒3c
2
−12c+11=
4
12
⇒3c
2
−12c+11=3
⇒3c
2
−12c+8=0
c=
2×3
12±
144−4×3×8
=
2×3
12±4
3
=
3
6±2
3
c=2−
3
2
3
=2−1.1547=0.8453
c=2−
3
2
3
=2+1.154=3.1543
∴c=
3
6±2
3
∴c∈(0,4)
Hence Lagrange's Mean Value theorem is verified.
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