Let $P_2(\mathbb R)$ be an inner product space with $\langle f, h \rangle = \int_{0}^{1}f(t)h(t)dt$. Let $g(f) = f(0) + f'(1)$. Find $h(t)$ such that $g(f) = \langle f,h \rangle$.
I tried representing $f$ as $f(t) = at^2+bt +c$ then $f(0) = c, f'(1) = 2a + b$. But I'm not sure if these are the right steps, and if they are, I don't know where to go from here.
It suffices to find $h \in P_2(\mathbb{R})$ such that $$g(f_i) = \langle f_i,h \rangle \text{ for } i = 0,1,2$$ where $f_i(t) = t^i, \beta = \{f_0,f_1,f_2\}$ is the standard basis for $P_2(\mathbb{R})$, as for the rest of $f \in P_2(\mathbb{R})$ the statement follows from the linearity of $g$ and $\langle\cdot,h\rangle$.
Let $h(t) = c_0 + c_1t + c_2t^2$. Then we have $$g(f_0) = \langle f_0,h\rangle \implies 1 = c_0 + \frac{1}{2}c_1 + \frac{1}{3}c_2$$ $$g(f_1) = \langle f_1,h\rangle \implies 1 = \frac{1}{2} c_0 + \frac{1}{3}c_1 + \frac{1}{4}c_2$$ $$g(f_2) = \langle f_2,h\rangle \implies 2 = \frac{1}{3}c_0 + \frac{1}{4}c_1 + \frac{1}{5}c_2$$
Solving $$\begin{pmatrix}1 & \frac{1}{2} & \frac{1}{3}\\\frac{1}{2} & \frac{1}{3} & \frac{1}{4}\\\frac{1}{3} & \frac{1}{4} & \frac{1}{5}\end{pmatrix} \begin{pmatrix}c_0\\c_1\\c_2\end{pmatrix} = \begin{pmatrix}1\\1\\2\end{pmatrix}$$ gives $$c_0 = 33, c_1 = -204, c_2 = 210$$ and finally $$h(t) = 33 - 204t + 210t^2$$