Optimization (1D min/max finder) implementation

Binding contract

This record describes what an executable surface must bind and validate. It does not claim that a public calculator, package, or production route is deployed.

Record owner

Guth Instruments

Lifecycle

verified-clean

Evidence ID

t8-024

Effect

Computation or source lookup only

Input and validation requirements

The specification exposes invalid-input hazards: a zero-width interval and an interval containing a pole must be rejected. It also requires explicit handling of nondifferentiable branch selection and angle-unit conventions; otherwise the returned extrema can be absent, undefined, or numerically wrong.

Required corrections

No binding erratum is attached to this unique-name record. Implementers must still preserve the declared units, domain, and failure behavior.

Inspectable interface fixture

The sample below is evidence for interface and validation tests, not a production implementation.

{
  "inputs": {
    "function": "f(x)=(x-2)^2+1",
    "search_interval": [
      0,
      4
    ],
    "method": "derivative_root_finding"
  },
  "expected_outputs": {
    "derivative": "f'(x)=2x-4",
    "stationary_points": [
      {
        "x": "2",
        "f_x": "1",
        "classification": "local and global minimum",
        "second_derivative": "2"
      }
    ],
    "endpoints": [
      {
        "x": "0",
        "f_x": "5"
      },
      {
        "x": "4",
        "f_x": "5"
      }
    ],
    "spot_checks": [
      {
        "at": 0,
        "input_side": 5,
        "answer_side": 5
      },
      {
        "at": 1.25,
        "input_side": 2.5625,
        "answer_side": 2.5625
      },
      {
        "at": 3.5,
        "input_side": 3.25,
        "answer_side": 3.25
      }
    ]
  },
  "reference_data": {
    "canonical_result": "f'(x)=0 gives 2x-4=0, hence x=2; f''(x)=2>0, so x=2 is a minimum; f(2)=1."
  },
  "working": "1. Differentiate: f'(x)=2(x-2)=2x-4. 2. Set the derivative to zero: 2x-4=0, so 2x=4 and x=2. 3. Evaluate the second derivative: f''(x)=2, and 2>0, so the stationary point is a local minimum. 4. Evaluate the function: f(2)=(2-2)^2+1=0+1=1. 5. Check endpoints: f(0)=(0-2)^2+1=4+1=5; f(4)=(4-2)^2+1=4+1=5. Therefore the reported minimum is (2,1)."
}

Other records

Sources and evidence

These pointers bind the page to evidence. They do not grant authority or collapse verification into deployment.

• VERIFIED_LEDGER.json
• evidence-tranche8/tranche8-results.json