Answer to Question #142438 in Calculus for Christian Selma

Question

Use the mean value theorem to prove 2.071 < squareroot of 4.3 < 2.075

Answer 1

Lemma :(mean value theorem)

if a function f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point c in the interval (a,b) such that f'(c) is equal to the function's average rate of change over [a,b].

Let f(x)=x³/3.

Then it is continuous on [2.071,2.075].

And differentiable on (2.071 , 2.075).

Then by mean value theorem there exists a point c in (2.071 , 2.075) such that f'(c)=(((2.075)³/3)-((2.071)³/3))/(2.075-2.071)

or, c²=4.3

Since c belongs to (2.071 , 2.075), therefore 2.071 < $\sqrt{4.3}$ < 2.075 (proved)