ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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For every $k\ge0$ , $x^k|x|$ is $C^k$ but not $C^{k+1}$

Example

For $k\in\mathbb N$ , the function $f_k(x)=x^k|x|$ is $C^k$ on $\mathbb R$ but not $C^{k+1}$ .

Facts & Assumptions

Given: $f_k(x)=x^k|x|$ .

[L1]

Absolute value is piecewise $x$ and $-x$ (Absolute value in an ordered field), and powers differentiate by For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term.

Verification

technique · cases

1.1

On $x>0$ , $f_k=x^{k+1}$ ; on $x<0$ , $f_k=-x^{k+1}$ .

assume-case positiveL1

2.1

Differentiating $j\le k$ times gives constant multiples of $x^{k+1-j}$ with opposite signs, and both one-sided values tend to $0$ . Defining the derivative value at $0$ by the difference quotient gives matching continuous derivatives through order $k$ .

step 1.1assume-case throughkL1

2.2

The $(k+1)$ -st one-sided derivatives are $(k+1)!$ and $-(k+1)!$ , so that derivative does not exist at $0$ .

assume-case nextorderstep 1.1L1

3.1

Hence $f_k\in C^k\setminus C^{k+1}$ according to Higher derivatives and the classes $C^k$ and $C^\infty$ .

step 2.1step 2.2cases-exhaustive∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 62 results over 17 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 lecture notes (standard reference, not scraped)

For every k≥0k\ge0k≥0, xk∣x∣x^k|x|xk∣x∣ is CkC^kCk but not Ck+1C^{k+1}Ck+1