The Inverse Function Theorem

Prove the 1D $C^1$ inverse function theorem completely: if $f$ is $C^1$ near $x_0$ and $f'(x_0) \ne 0$ , then $f$ is strictly monotone on some neighborhood of $x_0$ , hence has a continuous local inverse $g$ , and $g'(f(x_0)) = \frac{1}{f'(x_0)}$ .

f \in C^1,\ f'(x_0) \ne 0 \ \Rightarrow\ g'(y_0) = \frac{1}{f'(x_0)}, \quad y_0 = f(x_0)

f(x) = x^3+x

,

a = 1

:

f'(1) = 4

,

g'(2) = \frac{1}{4}

, product exactly 1 — as the theorem promises.

Show proof

Prove the case $f'(x_0) > 0$ (apply to $-f$ for the negative case). Since $f'$ is continuous and $f'(x_0) > 0$ , some neighborhood $I$ of $x_0$ has $f'(x) > 0$ throughout (sign preservation). By the MVT, $x_1 < x_2$ in $I$ gives $f(x_2) - f(x_1) = f'(\xi)(x_2 - x_1) > 0$ , so $f$ is strictly increasing on $I$ — hence injective, and by continuity it has a continuous inverse $g: J \to I$ on $J = f(I)$ .
Write $y_0 = f(x_0)$ , $y \in J$ , $x = g(y)$ , $x_0 = g(y_0)$ . The difference quotient is $\frac{g(y) - g(y_0)}{y - y_0} = \frac{x - x_0}{f(x) - f(x_0)}$ (for $x \ne x_0$ ; injectivity of $g$ gives $x \ne x_0$ when $y \ne y_0$ ). Applying the MVT to the denominator gives $\xi$ between $x$ and $x_0$ with $f(x) - f(x_0) = f'(\xi)(x - x_0)$ , so the quotient equals $\frac{1}{f'(\xi)}$ .
As $y \to y_0$ , continuity of $g$ gives $x = g(y) \to x_0$ , so the squeezed $\xi \to x_0$ . By continuity of $f'$ , $\frac{1}{f'(\xi)} \to \frac{1}{f'(x_0)}$ (denominator nonzero) — i.e. $g'(y_0) = \frac{1}{f'(x_0)}$ . Complete. (Note: the general $\mathbb{R}^n$ version is only sketched at idea level here, not proved: one views the derivative as a linear map, and the inverse matrix becomes the derivative of the local inverse.)

Verify the theorem's formula numerically for $f(x) = x^3+x$ at $a = 1$ : compute $f'(1) = 4$ , compute $g'(2)$ numerically via the bisection inverse, and check $|4\,g'(2) - 1| < 10^{-9}$ .

f(1) = 2, \quad f'(1) = 4, \quad g'(2) = \frac{1}{f'(g(2))} = \frac{1}{4}, \quad \lvert 4\,g'(2) - 1\rvert = 0 < 10^{-9}

a = 1

:

b = f(1) = 2

; bisection gives

g(2) = 1

, and

g'(2) = \frac{1}{f'(1)} = \frac{1}{4}

— check the product

4 \cdot \frac{1}{4} = 1

.

Show proof

$f'(x) = 3x^2+1$ , so $f'(1) = 4$ . With $b = f(1) = 2$ , bisection on $[-2, 4]$ converges to $g(2) = 1$ since $f$ is strictly increasing ( $f' \ge 1$ ); after 200 iterations the error is below $6/2^{200}$ , i.e. exactly 1 in double precision.
The formula $g'(y) = \frac{1}{f'(g(y))}$ at $y = 2$ gives $g'(2) = \frac{1}{f'(g(2))} = \frac{1}{f'(1)} = \frac{1}{3\cdot 1^2+1} = \frac{1}{4}$ . Hence $|4\,g'(2) - 1| = |4\cdot\frac{1}{4} - 1| = 0 < 10^{-9}$ — the theorem holds with no numerical error at all.

Show why the theorem breaks for $f(x) = x^3$ at $0$ : $f'(0) = 0$ , and the inverse $\sqrt[3]{y}$ is not differentiable at $0$ — prove this via the difference quotient blowing up to $\infty$ .

f'(0) = 0, \qquad \frac{\sqrt[3]{h} - 0}{h - 0} = h^{-2/3} \xrightarrow[h \to 0]{} \infty

The scene's

f(x) = x^3+x

satisfies the theorem even at

a = 0

since

f'(0) = 1 \ne 0

— the problem below studies

x^3

, where the derivative vanishes, showing why the hypothesis

f'(x_0) \ne 0

is needed.

Show proof

$f(x) = x^3$ is strictly increasing on all of $\mathbb{R}$ , so it has the global inverse $g(y) = \sqrt[3]{y}$ . But $f'(x) = 3x^2$ gives $f'(0) = 0$ — the theorem's hypothesis fails.
The difference quotient at $0$ : $\frac{g(h) - g(0)}{h} = \frac{\sqrt[3]{h}}{h} = h^{-2/3} = \frac{1}{(\sqrt[3]{h})^2}$ . As $h \to 0$ the denominator $\to 0^+$ , so the quotient blows up to $+\infty$ — no limit exists, hence $g$ is not differentiable at $0$ . The theorem's formula $\frac{1}{f'(0)}$ is likewise meaningless as $\frac{1}{0}$ .
Geometric meaning: when the graph of $f$ has a horizontal tangent at $(0,0)$ , the reflected graph of $g$ has a vertical tangent at $(0,0)$ — the failure of differentiability is visible in the picture too.

The Inverse Function Theorem

Examples & Exercises