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

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The function we have is f(x) = 2/e^(-2x) - 3/e^x

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

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

f'(x) = 2*2*e^2x - 3*(-1)*e^(-x)

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

f''(x) = 8*e^2x - 3*e^(-x)

2*f(x) + f'(x)

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

=> 4*e^2x - 6*e^(-x) + 4*e^2x + 3*e^(-x)

=> 8*e^2x - 3*e^(-x)

=> f''(x)

**This proves that 2*f(x) + f'(x) = f''(x) for 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).**