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Question

Verify Lagrange's Mean Value Theorem (LMVT) for following functions on indicated intervals. Also, find a point c in the indicated interval that satisfy LMVT.
i) f(x)=x(x2) on [1,3]
ii) f(x)=x(x1)(x3) on [0,1]

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Solution

LMVT states that,
If g:[a,b]R is a continuous function on [a,b] and differentiable on (a,b), then there exists some c in (a,b) such that g(c)=g(b)g(a)ba

i) f(x)=x(x2) on [1,3]
We know that, a polynomial function is everywhere continuous and differentiable. So, f(x) being a polynomial, is continuous on [1,3] and differentiable on (1,3).


Thus, f(x) satisfies both the conditions of LMVT on [1,3].
So, there must exist at least one real number c(1,3) such that f(c)=f(b)f(a)ba=f(3)f(1)31
f(x)=2x2,f(c)=2c2,f(3)=96=3 and f(1)=12=1
f(c)=f(3)f(1)312c2=3(1)312c2=2c=2
Thus, c=2(1,3) such that f(c)=f(3)f(1)31
LMVT is verified.

ii) f(x)=x(x1)(x3) on [0,1]
Since, a polynomial function is everywhere continuous and differentiable. So, f(x) being a polynomial, is continuous on [0,1] and differentiable on (0,1). Thus, f(x) satisfies both the conditions of LMVT on [0,1].
So, there must exist at least one real number c(0,1) such that f(c)=f(b)f(a)ba=f(1)f(0)10
f(x)=3x28x+3,f(c)=3c28c+3,f(1)=0 and f(0)=0.
3c28c+3=0010=0c=8±64366=8±286
c=473=0.451(0,1) such that f(c)=f(1)f(0)10
LMVT is verified.

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