To verify the identity, we'll have to calculate the first
and the 2nd derivatives of the fucntion f(x).
f'(x) =
[(3x+1)*e^-x]' = 3e^-x - (3x+1)*e^-x
f"(x) = [f'(x)]' =
[3e^-x - (3x+1)*e^-x]'
f"(x) = -3e^-x - 3e^-x +
(3x+1)*e^-x
Now, we'll substitute the expressions of f'(x)
and f"(x) into the given
identity:
f(x)=-2f'(x)-f"(x)
(3x+1)*e^-x
= -2[3e^-x - (3x+1)*e^-x] - [ -3e^-x - 3e^-x +
(3x+1)*e^-x]
We'll remove the
brackets:
(3x+1)*e^-x = -6e^-x + 2(3x+1)*e^-x + 3e^-x
+ 3e^-x - (3x+1)*e^-x
We'll combine and eliminate like
terms:
(3x+1)*e^-x = 2(3x+1)*e^-x -
(3x+1)*e^-x
(3x+1)*e^-x = (3x+1)*e^-x
q.e.d.
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