Show the total energy is conserved 
If the Lagrangian does not depend explicitly on time, then the quantity $E$ given by
  $$E := p\dot{x} - L \tag{1}$$
  is conserved.

I'm really confused. Normally the total energy is given by $$E = T + V.\tag{2}$$ Our definition of the $\textbf{Lagrangian}$  is
$$L(x,\dot{x}, t) = T - V\tag{3}$$
with $T$ being the kinetic energy and $V$ being the potential energy. So I think to rearrange to get
$$L = p\dot{x} - E = p\dot{x} - T - V.\tag{4}$$
But I don't know what the kinetic and potential are?
 A: I'm not the special case $L=T-V$ with $T=\frac{p^2}{2m},\,p=m\dot{x}$, you can show $p\dot{x}=2T$. The expression you've been given for $E$ is always conserved; you can prove this using an Euler-Lagrange equation.
A: *

*The precise statement is the following:



Given
  
  
*
  
*the definition of Lagrangian momentum $p_i:=\frac{\partial L}{\partial v^i}$,
  
*the definition of Lagrangian energy $h:=\sum_{i=1}^n p_i v^i-L $, and  
  
*the Euler-Lagrange eqs. $\frac{dp_i}{dt}\approx \frac{\partial L}{\partial q^i}$, 
  
*the Lagrangian $L(q,v)$ does not depend explicitly on time $t$,  
then the Lagrangian energy is conserved $\frac{dh}{dt}\approx 0$.



*

*For the perspective of Noether's theorem, see e.g. this related Phys.SE question.


Note that:


*

*A Lagrangian $L$ is not necessarily on the form $T-V$, cf. this Phys.SE post. 

*Mechanical energy $T+V$ is not necessarily the same as Lagrangian energy. 

*OP's Eqs. (2)-(4) do not necessarily hold. 
References:


*

*H. Goldstein, Classical Mechanics; Chapter 2.

