Hey, please tell me if the following is correct.
We have a continuous, increasing and strictly monotonic function on [MATH][a, b][/MATH], and [MATH]x_0\in[a, b][/MATH]. Let [MATH]f(y)[/MATH] be its inverse, and [MATH]f(x_0)=y_0[/MATH].
I want to show that [MATH]|y-y_0|<\delta\implies|g(y)-g(y_0)|<\epsilon[/MATH].
[MATH] |g(y)-g(y_0)| < \epsilon \Leftrightarrow x_0-\epsilon<g(y)<x_0+\epsilon\\ \Leftrightarrow f(x_0-\epsilon)<y<f(x_0+\epsilon)\\ \Leftrightarrow -(y_0-f(x_0-\epsilon))<y-y_0<f(x_0+\epsilon)-y_0 [/MATH]
If I let [MATH]\delta=\min(y_0-f(x_0-\epsilon),f(x_0+\epsilon)-y_0)[/MATH], while considering small [MATH]y[/MATH]s, then I think that I have it right.
For decreasing [MATH]f(x)[/MATH], I can flip the inequality symbols at the second step, and choose [MATH]\delta=\min(f(x_0-\epsilon)-y_0,y_0-f(x_0+\epsilon))[/MATH].
Thanks!
We have a continuous, increasing and strictly monotonic function on [MATH][a, b][/MATH], and [MATH]x_0\in[a, b][/MATH]. Let [MATH]f(y)[/MATH] be its inverse, and [MATH]f(x_0)=y_0[/MATH].
I want to show that [MATH]|y-y_0|<\delta\implies|g(y)-g(y_0)|<\epsilon[/MATH].
[MATH] |g(y)-g(y_0)| < \epsilon \Leftrightarrow x_0-\epsilon<g(y)<x_0+\epsilon\\ \Leftrightarrow f(x_0-\epsilon)<y<f(x_0+\epsilon)\\ \Leftrightarrow -(y_0-f(x_0-\epsilon))<y-y_0<f(x_0+\epsilon)-y_0 [/MATH]
If I let [MATH]\delta=\min(y_0-f(x_0-\epsilon),f(x_0+\epsilon)-y_0)[/MATH], while considering small [MATH]y[/MATH]s, then I think that I have it right.
For decreasing [MATH]f(x)[/MATH], I can flip the inequality symbols at the second step, and choose [MATH]\delta=\min(f(x_0-\epsilon)-y_0,y_0-f(x_0+\epsilon))[/MATH].
Thanks!
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