Exercise: G.M.V.T. use generalized mean value theorem 1

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Exercise: G.M.V.T. use generalized mean value theorem  1 < (e^(x)-1)/(ln(x+1)) < (x+1)e^(x)

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Understand GMVT: Let's first understand the Generalized Mean Value Theorem (GMVT). The GMVT states that if functions f and g are continuous on [a, b] and differentiable on (a, b), then there exists some c in (a, b) such that (f(b) - f(a)) / (g(b) - g(a)) = f'(c) / g'(c). We will apply this theorem to the functions f(x) = e^x - 1 and g(x) = ln(x + 1) on the interval [0, x] for x > 0.Check Functions: First, we need to check that f(x) = e^x - 1 and g(x) = ln(x + 1) are continuous on [0, x] and differentiable on (0, x). The function e^x is continuous and differentiable everywhere, and so is e^x - 1. The function ln(x + 1) is continuous and differentiable for x > -1, which includes our interval [0, x] since x > 0. Therefore, both f and g satisfy the conditions for the GMVT.Find Derivatives: Now, we will find the derivatives of f and g. The derivative of f(x) = e^x - 1 is f'(x) = e^x. The derivative of g(x) = ln(x + 1) is g'(x) = 1 / (x + 1).Apply GMVT: Applying the GMVT, there exists some c in (0, x) such that (f(x) - f(0)) / (g(x) - g(0)) = f'(c) / g'(c). Substituting the functions and their derivatives, we get: (e^x - 1 - (e^0 - 1)) / (ln(x + 1) - ln(0 + 1)) = e^c / (1 / (c + 1)). Simplifying, we have: (e^x - 1) / ln(x + 1) = e^c * (c + 1).Analyze Inequalities: Since c is in (0, x), we know that 0 < c < x. Therefore, e^c < e^x because the exponential function is increasing. Also, c + 1 < x + 1 for the same reason. Multiplying these two inequalities, we get e^c * (c + 1) < e^x * (x + 1).Show Positivity: From the previous step, we have (e^x - 1) / ln(x + 1) = e^c * (c + 1). Since e^c * (c + 1) < e^x * (x + 1), we can say that (e^x - 1) / ln(x + 1) < e^x * (x + 1).Consider Function h(x): Now we need to show that 1 < (e^x - 1) / ln(x + 1). We know that e^x is always greater than 1 for x > 0, so e^x - 1 is positive. Since ln(x + 1) is also positive for x > 0, the fraction (e^x - 1) / ln(x + 1) is positive. We need to show that it is greater than 1.Use Derivative of h(x): Consider the function h(x) = e^x - x - 1. Its derivative is h'(x) = e^x - 1. Since e^x > 1 for x > 0, h'(x) > 0, which means h(x) is increasing for x > 0. Since h(0) = 0, for x > 0, we have h(x) > 0, which means e^x - x - 1 > 0, or e^x - 1 > x.Combine Results: Since e^x - 1 > x, we can divide both sides by ln(x + 1) (which is positive for x > 0) to get (e^x - 1) / ln(x + 1) > x / ln(x + 1). Since the function x / ln(x + 1) approaches 1 as x approaches 0 from the right, and is increasing for x > 0, we have x / ln(x + 1) > 1 for x > 0. Therefore, (e^x - 1) / ln(x + 1) > 1.Combining the results from the previous steps, we have shown that 1 < (e^x - 1) / ln(x + 1) < e^x * (x + 1) for x > 0, which is what we wanted to prove using the Generalized Mean Value Theorem.

Exercise: G.M.V.T. use generalized mean value theorem $\newline$ 1<\frac{e^{x}-1}{\ln (x+1)}<(x+1) e^{x}

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Exercise: G.M.V.T. use generalized mean value theorem\newline 1<\frac{e^{x}-1}{\ln (x+1)}<(x+1) e^{x}

Exercise: G.M.V.T. use generalized mean value theorem $\newline$ 1<\frac{e^{x}-1}{\ln (x+1)}<(x+1) e^{x}