Inverse function theorem (one dimension). Let IRI\subseteq\mathbb R be an open interval and let f:IRf:I\to\mathbb R be continuously differentiable. If x0Ix_0\in I and f(x0)0f'(x_0)\ne0, then there are open intervals JIJ\subseteq I and KRK\subseteq\mathbb R, containing x0x_0 and f(x0)f(x_0), respectively, such that:

  1. The restriction fJ:JKf|_J:J\to K is bijective.
  2. Its g:KJg:K\to J is continuously differentiable and satisfies
    g(y)=1f(g(y)).g'(y)=\frac{1}{f'(g(y))}.
Remarks

In particular, g(f(x0))=1/f(x0)g'(f(x_0))=1/f'(x_0). Continuity of ff' makes ff' have constant sign near x0x_0; the derivative formula follows from the applied to fg=idf\circ g=\operatorname{id}.