Verify if 2f(x)+f'(x)= f''(x) f(x)=2/e^-2x-3/e^x.

To prove the given identity, first we need to determine
the expressions of the functions f(x),f'(x),f"(x), that represent the terms in
identity.

We'll re-write f(x) = 2/(1/e^2x) -
3*e^-x

f(x) = 2e^2x -
3*e^-x

We'll differentiate the function to get the 1st
derivative, f'(x):

f'(x) =
(2e^2x-3e^-x)'

f'(x) = 4e^2x -
(-3e^-x)

f'(x) = 4e^2x +
3e^-x

We'll differentiate now the expresison of the 1st
derivative, to get the 2nd derivative:

f"(x) =
[f'(x)]'

f"(x) = (4e^2x +
3e^-x)'

f"(x) = 8e^2x -
3e^-x

Now, we'll substitute the expressions of f"(x)
andf'(x) into the identity that has to be verified:

2(2e^2x
- 3*e^-x) + 4e^2x + 3e^-x = 8e^2x - 3e^-x

We'll remove the
brackets and we'll combine like terms:

4e^2x - 6e^-x+ 4e^2x
+ 3e^-x = 8e^2x - 3e^-x

LHS = 8e^2x - 3e^-x = 8e^2x - 3e^-x
= RHS

Since the LHS = RHS, the
identity is verified: 2f(x)+f'(x)= f''(x).