$$ % Define your custom commands here \newcommand{\bmat}[1]{\begin{bmatrix}#1\end{bmatrix}} \newcommand{\E}{\mathbb{E}} \newcommand{\P}{\mathbb{P}} \newcommand{\S}{\mathbb{S}} \newcommand{\R}{\mathbb{R}} \newcommand{\S}{\mathbb{S}} \newcommand{\norm}[2]{\|{#1}\|_{{}_{#2}}} \newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\pdd}[2]{\frac{\partial^2 #1}{\partial #2^2}} \newcommand{\vectornorm}[1]{\left|\left|#1\right|\right|} \newcommand{\abs}[1]{\left|{#1}\right|} \newcommand{\mbf}[1]{\mathbf{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\bm}[1]{\boldsymbol{#1}} \newcommand{\nicefrac}[2]{{}^{#1}\!/_{\!#2}} \newcommand{\argmin}{\operatorname*{arg\,min}} \newcommand{\argmax}{\operatorname*{arg\,max}} \newcommand{\dd}{\operatorname{d}\!} $$

Functional Equality

Differentiation

Problem 1 (Functional Equality) Let \(f: \R \to \R\) be a function such that \[ f(x+y) = f(x) + f(y) + 3xy. \tag{1}\] If \(f^\prime(1) = 4\), then find \(f^\prime(0)\).

Answer 1 Take the partial derivative of both sides of Equation 1 with respect to \(y\): \[ f^\prime(x + y) \cdot 1 = f^\prime(y) + 3x \] Evaluating this at \(y=0\) yields \(f^\prime(x) = f^\prime(0) + 3x\). Now, evaluate at \(x=1\) to obtain \[ 4 = f^\prime(1) = f^\prime(0) + 3. \] Solving for \(f^\prime(0)\), we find that \(\boxed{f^\prime(0) = 1}\).

As a bonus, along the way, we determined that \(f^\prime(x) = 1 + 3x\). This means that \[ f(x) = \int_0^x f^\prime(y) \dd y = \int_0^x (1 + 3y) \dd y = \frac{3}{2}x^2 + x + C. \] By Equation 1, we know that \(f(0) = 0\), which implies that \(C = 0\). Hence, we deduce that \[ \boxed{f(x) = x \left( 1 + \frac{3}{2}x \right)}. \]

← Back to Math