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How do I prove that (x-1)/x < lnx < (x–1), for x > 1 by using the mean value teorem?
The Mean Value Theorem is a fundamental theorem in calculus that relates the behavior of a function to the behavior of its derivative. It states that if a function is continuous on a closed interval [a,b] and differentiable on the open interval (a,b), then there exists at least one point c in (a,b) where the slope of the tangent line at c is equal to the average rate of change of the function over the interval [a,b].
The Mean Value Theorem is applied by considering the function f(x) = lnx - (x-1)/x. This function is continuous and differentiable on the interval [1,x] and satisfies the conditions of the Mean Value Theorem. By applying the Mean Value Theorem, we can show that there exists a point c in (1,x) where the slope of the tangent line is equal to the average rate of change of the function. This can be used to prove that (x-1)/x < lnx < (x-1) for all x > 1.
The Mean Value Theorem provides a way to connect the behavior of a function to the behavior of its derivative. In this case, we are able to use it to show that the function f(x) = lnx - (x-1)/x is strictly increasing on the interval [1,x]. This is crucial in proving the inequality (x-1)/x < lnx < (x-1), as it allows us to show that the function is always increasing and thus, the inequalities hold for all values of x > 1.
The Mean Value Theorem requires the function to be continuous on a closed interval and differentiable on the open interval. In this proof, we assume that the function f(x) = lnx - (x-1)/x satisfies these conditions on the interval [1,x]. Additionally, we assume that x > 1, as the inequality does not hold for x ≤ 1.
Yes, the Mean Value Theorem can be applied to prove other inequalities involving functions. It is a versatile tool in calculus and is often used in proofs related to the behavior of functions and their derivatives. It is also used in other areas of mathematics, such as optimization problems and differential equations.