I want to maximize the following function: $$f(\theta)=\frac{1}{\sqrt{2\pi c\theta}}e^{-\frac{1}{2c\theta}(x-\theta)^2}$$
with respect to $\theta\in[0,1]$. Assume the rest of the variables known and $c>0$. Therefore, I took the log: \begin{eqnarray*} \mathrm{ln}(f(\theta))&=&\mathrm{ln}(2\pi c\theta)^{-1/2}-\frac{1}{2c\theta}(x-\theta)^2\\ &=&-\frac{1}{2}\mathrm{ln}(2\pi c\theta)-\frac{x^2}{2c\theta}+\frac{x}{c}-\frac{\theta}{2c} \end{eqnarray*} and eventually \begin{eqnarray*} \frac{\vartheta\mathrm{ln}(f(\theta))}{\vartheta\theta}=0&\Rightarrow&-\frac{1}{2}\frac{2\pi c}{2\pi c\theta}+\frac{x^2}{2c}\frac{1}{\theta^2}-\frac{1}{2c}=0\\ &\Rightarrow&\frac{x^2-c\theta}{c\theta^2}=\frac{1}{c}\\ &\Rightarrow&\theta^2+c\theta-x^2=0 \end{eqnarray*} Is this a correct approach? What should I do next? I also need to satisfy the constraint $\theta\in[0,1]$.
Hint. What you have done is right. Then solving the quadratic equation gives two potential solutions $$ \theta_0=-\frac12 \left(\sqrt{c^2+4 x^2}+c\right), \qquad \theta_1=\frac12 \left(\sqrt{c^2+4 x^2}-c\right). $$ Which $\theta_i$ satisfies $0\le\theta_i \le1$ ?