Inverse function theorem (one dimension). Let I⊆RI\subseteq\mathbb R be an open interval and let f:I→Rf:I\to\mathbb R be continuously differentiable. If x0∈Ix_0\in I and f′(x0)≠0f'(x_0)\ne0, then there are open intervals J⊆IJ\subseteq I and K⊆RK\subseteq\mathbb R, containing x0x_0 and f(x0)f(x_0), respectively, such that:

  1. The restriction f∣J:J→Kf|_J:J\to K is bijective.
  2. Its g:K→Jg: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 f′f' makes f′f' have constant sign near x0x_0; the derivative formula follows from the applied to f∘g=id⁡f\circ g=\operatorname{id}.