= Solution
Take the affine trial space $V_1=\{v\in H^1(a,b):v(a)=1\}$ and the test space $V_0=\{v\in H^1(a,b):v(a)=0\}$. Multiplication by $v\in V_0$ and <integration by parts> incorporate the right <Neumann boundary condition> as a load:
$$
\boxed{u\in V_1,\qquad\int_a^bu'v'\,dx=\int_a^bfv\,dx+2v(b)\quad(v\in V_0).}
$$
For the continuous piecewise-quadratic trial space, impose the left value exactly and use the same equation for every piecewise-quadratic test function vanishing at $a$. With a <basis> $(\psi_j)$ of that test space and $u_h=1+\sum_jc_j\psi_j$, the <finite element> equations are
$$
\boxed{\sum_j\left(\int_a^b\psi_j'\psi_i'\,dx\right)c_j=\int_a^bf\psi_i\,dx+2\psi_i(b).}
$$
The <Endpoint Poincaré inequality>, reflected about the interval midpoint, makes the derivative norm a norm on $V_0$. <Galerkin orthogonality> then gives the exact best-approximation identity: for every admissible $v_h$,
$$
\|u'-v_h'\|_2^2=\|u'-u_h'\|_2^2+\|u_h'-v_h'\|_2^2.
$$
Therefore \b[$u_h$ minimizes the derivative error over all admissible piecewise-quadratic functions].
To prove the requested error bound, assume $f\in H^1(a,b)$. Then $u\in H^3(a,b)$ and $u'''=-f'$. We need an elementary <Poincare-Wirtinger inequality> on each element $I$ of length $h_I$: if $w\in H^1(I)$ has mean zero, its continuous representative has a zero somewhere, and the <fundamental theorem of calculus> and <Cauchy-Schwarz inequality> give $\|w\|_{L^2(I)}\leq h_I\|w'\|_{L^2(I)}$.
Choose an affine function $q_I$ whose slope is the average of $u''$ on $I$ and whose average equals the average of $u'$. Then $w=u'-q_I$ and $w'=u''-q_I'$ both have mean zero. Applying the just-proved bound twice yields
$$
\|u'-q_I\|_{L^2(I)}\leq h_I^2\|u'''\|_{L^2(I)}.
$$
Define a quadratic interpolant on $I$ by integrating $q_I$ from the true left-endpoint value. The equal-average condition ensures that it reaches the true right-endpoint value, so these pieces join to an admissible continuous piecewise-quadratic function $I_hu$. Sum the squared estimates and use the best-approximation property:
$$
\boxed{\|u'-u_h'\|_2\leq\|u'-(I_hu)'\|_2\leq h^2\|u'''\|_2=h^2\|f'\|_2.}
$$
Only the maximum element length is used; equal spacing is unnecessary.
For the <Aubin–Nitsche duality argument>, let $e=u-u_h$ and solve $-z''=e$, $z(a)=0$, $z'(b)=0$. Explicitly $z'(x)=\int_x^be(t)\,dt$, so $\|z''\|_2=\|e\|_2$. Integration by parts and <Galerkin orthogonality> give
$$
\|e\|_2^2=\int_a^be'z'\,dx=\int_a^be'[z'-(J_hz)']\,dx,
$$
where $J_hz$ is its nodal piecewise-linear interpolant, which belongs to the quadratic test space. On each element $(J_hz)'$ is the average of $z'$, so the same mean-zero inequality gives $\|z'-(J_hz)'\|_2\leq h\|z''\|_2$. Canceling $\|e\|_2$ when nonzero proves the additional order:
$$
\boxed{\|u-u_h\|_{L^2(a,b)}\leq h\|u'-u_h'\|_2\leq h^3\|f'\|_2.}
$$
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