Worked examples and diagnosis

A complete constrained calculation

A model C(x)=2x+18/x is valid only for 1≤x≤6. Differentiate: C′=2−18/x²; positive critical point x=3. Candidate values are C(1)=20, C(3)=12, C(6)=15. The continuous function attains a minimum on the closed interval; these include endpoints and all interior critical points, so 12 is the global minimum. Replacing the domain with [4,6] moves the optimum to x=4, value 12.5. An unchanged answer of 3 violates the new constraint.

Density, cumulative probability and expectation

Let f(t)=c t² for 0≤t≤3, zero elsewhere. Nonnegativity requires c≥0. Its integral is 9c, so c=1/9. F(t) is zero below 0, t³/27 on [0,3], and one above 3. Thus P(1≤T≤2)=7/27. The mean is ∫₀³ t³/9 dt=9/4. Check the mean lies in the support. Density at 3 is 1 per unit of t; this is not P(T=3), which is zero.

Separate numerical and tail errors

For density 2exp(−2t), approximate total mass by trapezoids on [0,5] with n subintervals. The exact finite integral is 1−exp(−10), so omitted mass is exp(−10), about 0.0000454. The second derivative is 8exp(−2t), bounded by 8. Quadrature error is at most 8·5³/(12n²). At n=300 this is about 0.000926; together with the tail bound it is below 0.001. This conservative guarantee ignores floating-point rounding and assumes the stated density model. A measured smaller error does not invalidate a conservative bound.

Diagnose a false root guarantee

The function 1/x has opposite signs at −1 and 1, but is not continuous on that interval and has no root. Bisection’s theorem therefore does not apply. For x²−2 on [1,2], continuity does apply; ten halvings produce a bracket of width 1/1024 and midpoint error at most 1/2048. A tolerance on function value alone would not give that positional guarantee for arbitrary functions.