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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(x−2) on [1,3] ii) f(x)=x(x−1)(x−3) 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)b−a i) f(x)=x(x−2) 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)b−a=f(3)−f(1)3−1 f′(x)=2x−2,f′(c)=2c−2,f(3)=9−6=3 and f(1)=1−2=−1 f′(c)=f(3)−f(1)3−1⇒2c−2=3−(−1)3−1⇒2c−2=2⇒c=2 Thus, c=2∈(1,3) such that f′(c)=f(3)−f(1)3−1 ⇒ LMVT is verified. ii) f(x)=x(x−1)(x−3) 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)b−a=f(1)−f(0)1−0 f′(x)=3x2−8x+3,f′(c)=3c2−8c+3,f(1)=0 and f(0)=0. ⇒3c2−8c+3=0−01−0=0⇒c=8±√64−366=8±√286 ⇒c=4−√73=0.451∈(0,1) such that f′(c)=f(1)−f(0)1−0 ⇒ LMVT is verified.

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