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How do you find the value of square root of 15 using lagrange's mean value theorem?
Wiki User
∙ 10y ago
The square root of 15 is an irrational number and to 9 decimal places it is about 3.872983346
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Let me write the Mean Value Theorem (mvt) as f'(c) = (f(b)-f(a))/(b-a), for some c in [a,b]. f is the square root function, which is obviously continuous (and diff'able) on [a,b], provided that a>=0. f'(c) = (1/2)c-1/2.
If I pretend to ignore the first answer, I can say that a little mental experimentation would indicate that f(15) is inside [3.844, 4]. And I chose these bounds because they give 'easy' square roots.
Applying the mvt, (1/2)c-1/2 = ( f(16)-f(15) ) / (16-15) = 4 - f(15) for some c in (3.844, 4).
(1/2)c-1/2 is a decreasing function, hence it has its lower bound at 4; therefore,
4 - f(15) > (1/2)4-1/2 = 1/4 => f(15) < 15/4
(1/2)c-1/2 has its upper bound at 3.844; hence,
4 - f(15) < (1/2)3.844-1/2
Put these together to obtain bounds for f(15).
Wiki User
∙ 10y ago
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