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Prove the inequalities 1 – 1/x < log x < x - 1

  1. Define the following functions:

        \[ f(x) = x - 1 - \log x, \qquad g(x) = \log x - 1 + \frac{1}{x}, \]

    for x > 0. Prove that for x > 0 and x \neq 1 the inequalities

        \[ 1 - \frac{1}{x} < \log x < x - 1 \]

    hold. Proceed by examining the signs of the derivatives f' and g'. When x = 1 these are equalities.

  2. Draw the graphs of the functions

        \[ A(x) = x - 1, \qquad B(x) = 1 - \frac{1}{x}, \qquad x > 0. \]

    Interpret the inequalities in part (a) geometrically.


  1. First, we compute the derivatives of the functions,

        \begin{align*}  f(x) &= x - 1 - \log x & \implies && f'(x) &= 1 - \frac{1}{x} \\  g(x) &= \log x - 1 + \frac{1}{x} & \implies && g'(x) &= \frac{1}{x} - \frac{1}{x^2} = \frac{1}{x} \left( 1 - \frac{1}{x} \right). \end{align*}

    Then, considering the derivative of f,

        \begin{align*}   f'(x) = 1 - \frac{1}{x} && \implies && f'(x) > 0 \quad &\text{when } x > 1 \\  \text{and} && \implies && f'(x) < 0 \quad &\text{when } 0 < x < 1. \end{align*}

    Therefore, the function f has a minimum at x = 1. Since we can directly evaluate f(1) = 0, this means f(x) > 0 for x > 0 and x \neq 1. Therefore, x - 1 > \log x.

    Next, looking at the derivative of g,

        \begin{align*}  g'(x) = \frac{1}{x} \left( 1 - \frac{1}{x} \right) && \implies && g'(x) > 0 \quad & \text{when } x > 1 \\  \text{and} && \implies && g'(x) < 0 \quad &\text{when } 0 < x < 1. \end{align*}

    Therefore, the function g has a minimum at x = 1. Since g(1) = 0 this implies g(x) > 0 for x > 0 and x \neq 1. Thus,

        \[ 1 - \frac{1}{x} < \log x. \]

    Putting these two pieces together we have established the requested inequalities:

        \[ 1 - \frac{1}{x} < \log x < x - 1 \qquad \text{for } x > 0 \text{ and } x \neq 1.  \qquad \blacksquare\]

  2. The graph of the two functions is:

    Rendered by QuickLaTeX.com

    The inequalities in part (a) imply that the graph of \log x must lie strictly between the graphs of A(x) and B(x) shown above (and so \log 1 = 0).

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