The Inverse Function Theorem

Near a point with nonzero slope, a function can be flipped. The flipped function's slope is the reciprocal of the original.

Flipping swaps horizontal and vertical. So the ratio of rise to run turns upside down. Multiply the two tangent slopes and you get exactly 1.

For example, imagine 💰 money and 🍬 candies traded at a 🏪 shop: 4 money for 1 candy. Traded back, 1 money buys a quarter of a candy — the rate flips into its reciprocal.

Examples & Exercises

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 interval $I$ 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.
Since $f$ is continuous, the intermediate value theorem (IVT) makes $J = f(I)$ an interval and $f: I \to J$ surjective. The inverse of a strictly monotone function is continuous: if $y_n \to y$ but $g(y_n) \not\to g(y)$ , the order is violated, contradicting monotonicity of $f$ (the standard $\varepsilon$ - $\delta$ /order argument).
Hence $f$ 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.