Find solution of maximum of a function with conditions 
\begin{equation} \label{eqgen}
\begin{aligned}
& \underset{x}{\text{maximize}}
& & F(x)\\
& \text{subject to}
& & x \geq 0, \\
& & &x + f(x)\leq d,
\end{aligned}
\end{equation}
  where $F(x), f(x)$ are continuous functions from $\mathbb{R}^+ \to \mathbb{R}^+$.

Here is my approach. Intuitively, $x+f(x)\leq d, x\geq 0$ is equivalent to $x \in [a_1, b_1] \cup [a_2, b_2]\cup \cdots \cup [a_{n}, b_{n}]$,
where $a_i\leq b_i$, $a_1$ is either 0 or solution of $x+f(x) = d$, $b_n$ is either d or solution of $x+f(x) = d$, and $a_i, b_i$ are solutions of $x+f(x) = d$ for other cases. 
Hence, the optimization problem becomes
\begin{equation} 
\begin{aligned}
& \underset{x}{\text{maximize}}
& & F(x)\\
& \text{subject to}
& & x \in  [a_1, b_1] \cup [a_2, b_2]\cup \cdots \cup [a_{n}, b_{n}].
\end{aligned}
\end{equation}
max $F(x)$ on $[a_i, b_i]$ is attained at either $a_i, b_i$ or maximum of $F(x)$ on $[a_i, b_i]$. We can conclude that the optimal solution is either solutions of $x+f(x) = d, or x=0, x= d$  or local maximum of function $F(x)$. This is also what I want to prove. 
How can I formally write down the solution of the above approach? I wrote this and my professor does not accept my solution. Thank you in advance!

 A: There are a lot of intuitive tricks that one can use if one knows exactly the values of $f(x),\  d$, but given it is not given, I will lay out a general way of Kuhn-Tucker conditions.
Write: $L=F(x)+\lambda_1 x-\lambda_2 (x+f(x)-d)  $ with $\lambda_1, \lambda_2 \geq0$
Think of the intuition behind the sign of the lambdas. If $x$ becomes negative that is bad for our purpose given the constraint, and that is getting reflected by our objective function getting a negative "shock" through the positive $\lambda_1$. You can reason through similarly for the other lambda.
Now, take derivative w.r.t $x$ to get:
$F'(x)-\lambda_1+\lambda_2(1+f'(x))=0$
Couple this equation with the constraints:
$\lambda_1 x=0$
$\lambda_2 (x+f(x)-d)=0$
And these conditions have complementary slackness.
Case 1: Suppose both constraints are binding, then $\lambda_1, \lambda_2>0$. And, $x=0$, and, $f(0)=d$. If, $f(0) \ne d$, we can immediately move on from this case to the second case. 
Otherwise, plug in $x=0$ in the derivative condition, solve for lambdas and check if they are actually positive, otherwise, as before, we move on from this case. Check here:
$F'(0)-\lambda_1+\lambda_2(1+f'(0))=0$
Case 2: Constraint 1 binds and 2 does not.
So, $\lambda_1>0$ and $\lambda_2=0$ and $x=0$.
Check back into the derivative condition and check if the values match up with it and with the condition
$ f(0)\leq d $
Case 3: Constraint 2 binds and 1 does not.
So, $\lambda_1=0$ and $\lambda_2>0$ and $x=0$.
Solve for $x$ from $f(x)=d$, given we are not in case 1, $x$ should not be zero. Use that value of the solution to check back if $x\geq 0$ holds, and if the derivative condition is satisfied for positive $\lambda_2$.
Case 4: Both constraints bind
So, $\lambda_1=0$ and $\lambda_2=0$ 
Solve it like an unconstrained problem, find $x$ and check back to confirm that the inequality conditions are satisfied.
Finally at any case when you find a solution, you check for the second derivative of the maximand (Lagrangian function L) to be negative at the optimal point ($x^*, \lambda^*$). Also, if you found multiple solutions from the different cases, choose among them by plugging them directly into the actual function $F$ to see which one gives the maximum value.
