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Question
Verify the Lagrange’s mean value theorem for the function:
`f(x)=x + 1/x ` in the interval [1, 3]
Solution
`f(x)=x + 1/x ` x ∈ [1, 3]
(i) f (x) is continuous for x ∈ [1,3]
(ii) f (x) is differentiable for x ∈ (1,3)
∴ LMVT is applicable.
`f(1)=1+1/1 = 2 and f(3)= 3 + 1/3 = 10/3 `
`f'(x)= 1- 1/x^2 therefore f'(c)= 1-1/c^2`
`therefore f'(c)= (f(b)-f(a))/(b - a) rArr 1-(1)/c^2= (10/3 -2)/(3-1) = ((10-6)/3)/2 = 4/6=2/3`
`rArr 1- 2/3 = 1/c^2 rArr 1/3=1/c^2 ∴ c^2 = 3`
∴ c = ± `sqrt3 but c=sqrt3 ∈ [1,3]`
LMVT is vertified.
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