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Question

Verify Langrange's mean value theorem for the function:
f(x)=x(1logx) and find the value of 'c' in the interval [1, 2].

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Solution

f(x)=x(1logx)
The function f(x) is continuous in [1,2] because x is continuous every where while logx is continuous when xϵ[1,]

The function f(x) is also differential in (1,2)

Therefore there exist c , such that c belongs to (1,2)
f(c)=f(b)f(a)ba

f(c)=f(2)f(1)21

f(c)=(22log2)(1log1)21

f(c)=12log221

f(x)=logx

f(c)=logc

logc=12log2

logc=2log21

logc=log41

log4logc=1

log4c=1

4c=e1

c=4e1

c=1.472

Thus the value of c belongs to (1,2)

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