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Show a function satisfies the hypotheses of the mean value theorem

by RoRi

Let

    \[ f(x) = \begin{dcases} \frac{3-x^2}{2} & \text{if } x \leq 1, \\ \frac{1}{x} & \text{if } x \geq 1. \end{dcases} \]

  1. Draw the graph of f for x \in [0,2].
  2. Show that the hypotheses of the mean value theorem are satisfied on [0,2] and find the mean values the theorem provides.
  1. The sketch is as follows:

    Rendered by QuickLaTeX.com

  2. Since \frac{3-x^2}{2} and \frac{1}{x} are continuous on [0,1] and [1,2], respectively and are differentiable on (0,1) and (1,2), the only point at which this piecewise function might be discontinuous or non differentiable is at x = 1.

    At x = 1 we have

        \[ \lim_{x \to 1^-} \frac{3-x^2}{2} = 1, \qquad \lim_{x \to 1^+} \frac{1}{x} = 1. \]

    Thus, the left and right-hand limits are equal, so the limit exists and equals the function value. Therefore, f is continuous at x = 1. Finally, we must check that the derivative exists at x = 1. Since the derivative of \frac{3-x^2}{2} = -x and the derivative of \frac{1}{x} = -\frac{1}{x^2}, both are equal to -1 at x = 1. Hence, the derivative exists.

    Now, we have met the conditions of the mean-value theorem, so we can apply the theorem to conclude,

        \begin{align*} f(2) - f(0) = f'(c)\cdot 2 &\implies \frac{1}{2} - \frac{3}{2} = 2 f'(c) \\ &\implies f'(c) = -\frac{1}{2}. \end{align*}

    Further, we know

        \[ f'(x) = \begin{dcases} -x & \text{if } x < 1 \\ -\frac{1}{x^2} & \text{if } x \geq 1. \end{dcases} \]

    So,

        \[ f'(c) = -c = -\frac{1}{2} \quad \implies \quad c = \frac{1}{2} \]

    and

        \[ f'(c) = -\frac{1}{c^2} = -\frac{1}{2} \quad \implies \quad c = \sqrt{2} \qquad (\text{since } -\sqrt{2} \notin [1,2].) \]

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