Problem 46 Prove the inequality $$ \fra... [FREE SOLUTION]

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Single Variable Calculus

Soo T. Tan

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Chapter 3: Problem 46

Prove the inequality $$ \frac{x}{x+1}<\ln (1+x)0\).

Short Answer

Expert verified

The inequality $\frac{x}{x+1}<\ln (1+x) 0$ can be proven by analyzing it in two parts: the left inequality $\frac{x}{x+1}<\ln (1+x)$ and the right inequality $\ln (1+x) 0$ and that they are equal to 0 when $x=0$. Thus, we can conclude that the given inequality is true for all $x > 0$.

Step by step solution

Left Inequality Proving \(\frac{x}{x+1}

To prove the left inequality $\frac{x}{x+1}<\ln (1+x)$, let's consider a function $f(x) = \ln(1+x) - \frac{x}{x+1}$. It is obviously defined for $x>0$. Our goal is to show that $f(x) > 0$ for all $x > 0$.

Find the derivative of f(x)

To analyze f(x), we will find its derivative with respect to x, $f'(x)$. By applying the chain rule and quotient rule, we get: $f'(x) = \frac{1}{1+x} - \frac{1(x+1) - x}{(x+1)^2} = \frac{1}{1+x} - \frac{1}{(x+1)^2}$

Simplifying the derivative

Now, let's simplify the derivative: $f'(x) = \frac{(x+1) - 1}{(x+1)^2} = \frac{x}{(x+1)^2}$ Since $x > 0$ and $(x+1)^2 > 0$, it's clear that the derivative $f'(x) > 0$ for all $x > 0$.

Using the monotonic property

Since $f'(x) > 0$, the function $f(x)$ is monotonically increasing for $x > 0$. Thus, we only need to check the limit as x approaches 0: $\lim_{x \to 0} f(x) = \lim_{x \to 0} \left[ \ln(1+x) - \frac{x}{x+1} \right] = \ln(1) - \frac{0}{1} = 0$ Since $f(x)$ is monotonically increasing and $f(0)= 0$, it follows that $f(x) > 0$ for all $x > 0$. Thus, the left inequality $\frac{x}{x+1}<\ln (1+x)$ is true for all $x > 0$.

Right Inequality Proving $\ln (1+x) < x$

To prove the right inequality $\ln (1+x) < x$, let's consider a function $g(x) = x - \ln (1+x)$. It is also defined for $x>0$. Our goal is to show that $g(x) > 0$ for all $x > 0$.

Find the derivative of g(x)

To analyze g(x), we will find its derivative with respect to x, $g'(x)$. By applying the chain rule, we get: $g'(x) = 1 - \frac{1}{1+x}$

Simplifying the derivative

Now, let's simplify the derivative: $g'(x) = \frac{x}{x+1}$ Since $x > 0$ and $x+1 > 0$, it's clear that the derivative $g'(x) > 0$ for all $x > 0$.

Using the monotonic property

Since $g'(x) > 0$, the function $g(x)$ is monotonically increasing for $x > 0$. Thus, we only need to check the limit as x approaches 0: $\lim_{x \to 0} g(x) = \lim_{x \to 0} \left[ x - \ln(1+x) \right] = 0 - \ln(1) = 0$ Since $g(x)$ is monotonically increasing and $g(0)= 0$, it follows that $g(x) > 0$ for all $x > 0$. Thus, the right inequality $\ln (1+x) < x$ is true for all $x > 0$.

Combine the inequalities

Finally, combining the results from steps 4 and 8, we have proven that $\frac{x}{x+1}<\ln (1+x) 0$.

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91影视

Prove the inequality $$ \frac{x}{x+1}<\ln (1+x)0\).

Short Answer

Step by step solution

Left Inequality Proving \(\frac{x}{x+1}

Find the derivative of f(x)

Simplifying the derivative

Using the monotonic property

Right Inequality Proving \(\ln (1+x) < x\)

Find the derivative of g(x)

Simplifying the derivative

Using the monotonic property

Combine the inequalities

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