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A linear interpolant is the unique polynomial of polynomial degree at most one taking prescribed values $y_a,y_b$ at distinct interpolation nodes $a,b$. It is an affine function, including the constant function case. For $a\le x\le b$, its two displayed coefficients are nonnegative and sum to one, so changing each sampled value by at most $\delta$ changes every interpolated value by at most $\delta$. Differentiation instead uses the difference of the two samples divided by $b-a$, causing noise amplification by differentiation of a linear interpolant.

For linear interpolation at zero and $h>0$, an error $\delta$ in the second sampled value, with the first exact, changes the linear interpolant by $\Delta p(x)=\delta x/h$. On $[0,h]$, its uniform norm is exactly $|\delta |$, whereas the derivative change is the constant $\delta /h$. With independent absolute bounds $|{\delta }_0|,|{\delta }_1|\le \delta$ on both samples, the value change remains at most $\delta$ and the derivative change is at most $2\delta /h$, attained for opposite endpoint errors.

Thus bounded absolute sample noise can remain bounded under value interpolation while becoming unbounded under numerical differentiation as $h\to 0$. Fixed nonzero sample noise also prevents exact convergence of the value approximation; stability against perturbations and convergence to exact data are different assertions. For a second-sample error $aϵ(h)$ with fixed $a>0$, value convergence requires $ϵ(h)\to 0$, while derivative convergence requires $ϵ(h)=o(h)$ in little o notation, in addition to vanishing truncation error.

The trapezoidal rule is exact for every affine function $f(x)=\alpha +\beta x$, equivalently every polynomial of polynomial degree at most one: the linear interpolant is the function itself. These functions have $f^{\prime \prime }=0$, so the preceding upper bound is also zero. There is no contradiction: zero error is precisely what the bound predicts.

The degree of precision of a quadrature rule is one, since the quadratic polynomial in part (i) is not integrated exactly. This does not exclude exactness for particular higher-degree or nonpolynomial functions. For example $f(x)=(x-m{)}^3$, $m=(a+b)/2$, has zero integral and opposite endpoint values, so the trapezoidal rule also gives zero. Its second derivative is generally nonzero, making the upper bound positive; an upper bound need not be attained. Signed quadrature error can cancel.

It suffices to assume $f\in C^2[a,b]$. The polynomial interpolation error for the endpoint linear interpolant gives, for every interior $x$, some ${\xi }_x\in (a,b)$ withLet $M={\max }_{[a,b]}|f^{\prime \prime }|$, which exists by continuity of the second derivative and the extreme value theorem. Since $(x-a)(b-x)\ge 0$,Integrating this pointwise inequality and using the triangle inequality for an integral yieldsThusThis derivation does not require choosing ${\xi }_x$ as a continuous function of $x$; the pointwise upper bound is sufficient.

The constant is sharp. Take the quadratic polynomial $f(x)=(x-a{)}^2$ and write $L=b-a$. Its exact integral is $L^3/3$, its trapezoidal rule value is $L^3/2$, and $\max |f^{\prime \prime }|=2$. ConsequentlyEquality holds, so no smaller universal constant than $1/12$ works under these assumptions. This is the sharp one-panel trapezoidal error bound.

For $f\in C^2[a,b]$ with $a<b$, the trapezoidal rule satisfies the displayed bound, and the constant $1/12$ is best possible. The polynomial interpolation error for the endpoint linear interpolant satisfies $|f(x)-p(x)|\le \frac{1}{2}M(x-a)(b-x)$, where $M=\max |f^{\prime \prime }|$. Integrating this inequality and using the triangle inequality for an integral proves the bound because ${\int }_a^b(x-a)(b-x)dx=(b-a{)}^3/6$. Equality holds for $f(x)=(x-a{)}^2$: the exact integral is $(b-a{)}^3/3$, its trapezoidal rule estimate is $(b-a{)}^3/2$, and $M=2$.