Write V_final in terms of V1
I learned to solve these diode-related problems like this.

First, I'll re-draw your schematic:







simulate this circuit &ndash; Schematic created using CircuitLab

Then, look at two cases separately:

1) D1 is off (not conducting), so the voltage across the diode is less than 0.7 V

No current goes through R1, so \$V_{out} = V_1\$. The voltage across the diode is 

\$ V_{D1} = -5V - V_{out} &lt; 0.7V\$

so

\$ V_1 &gt; -5.7 V\$

2) D1 is on (conducting). We know from case 1 that \$V_1 &lt; -5.7V\$.

Taking KVL around the circuit and guessing (correctly) that current moves counterclockwise, we get

\$V_1 + 5V + 0.7V + (20kΩ)I  = 0\$

so

\$I = - \frac{V_1 + 5.7V}{20kΩ}\$

and the output voltage is 

\$V_{out} = V_1 + IR_1 
= V_1 - (10kΩ)\frac{V_1 + 5.7V}{20kΩ}
= \frac{1}{2}V_1 - 2.85V\$

I simulated this circuit in LTSpice. Here is a plot of \$V_{1}(t)\$ and \$V_{out}(t)\$:


Notice that \$V_{out} = V_1\$ until \$V_1 &lt; 5.7V\$, where it gets attenuated by the voltage divider of the two resistors.