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Arcosine function

Given a function f:[-1, 1] \rightarrow [0, \pi] such that

(1)   \begin{equation*} \cos f(x) = x \quad \text{for all} \;\; x \in [-1 , 1] \end{equation*}

  1. Evaluate f(0).
  2. Prove that f is one to one.
  3. Prove that f \left ( \cos x \right ) = x for all x \in [0, \pi].
  4. Find the range of f.
  5. Sketch the graph of f.

Solution

  1. Setting x=0 at (1) we have that

        \[\cos f(0) =0 \Leftrightarrow x = \frac{\pi}{2} + \kappa \pi \; , \; \kappa \in \mathbb{Z}\]

    However,

        \begin{align*} 0 \leq f(0) \leq \pi &\Rightarrow 0 \leq \kappa \pi + \frac{\pi}{2} \leq \pi \\ &\Rightarrow - \frac{\pi}{2} \leq \kappa \pi \leq \frac{\pi}{2} \\ &\Rightarrow -\frac{1}{2} \leq \kappa \leq \frac{1}{2} \\ &\!\!\!\!\!\!\overset{\kappa \in \mathbb{Z}}{=\! =\! =\!\Rightarrow } \kappa =0 \end{align*}

    Thus f(0)=\frac{\pi}{2}.

  2. Let x_1, x_2 \in [-1, 1] such that f(x_1) = f(x_2). Thus,

        \begin{align*} f(x_1) = f(x_2) &\Leftrightarrow \cos f(x_1) = \cos f(x_2) \\ &\Leftrightarrow x_1 = x_2 \end{align*}

    Hence f is 1-1.

  3. Setting x = \cos x we have

        \begin{align*} \cos f(x) =x &\Leftrightarrow \cos f \left ( \cos x \right ) = \cos x \\ &\Leftrightarrow f \left ( \cos x \right ) = x \end{align*}

    since \cos is strictly decreasing in [0, \pi].

  4. f \left( [-1, 1] \right) = [ 0, \pi] because f^{-1}(x) = \cos x \; , \; x \in [0, \pi].
  5. The graph is seen at the following figure:

    Rendered by QuickLaTeX.com

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