Verify that f(x)=-2f'(x)-f"(x) if f(x)=(3x+1)*e^-x
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We have f(x) = (3x + 1)*e^-x
We use the product rule to find f'(x)
f'(x) = (3x + 1)'*e^-x + (3x + 1)*(e^-x)'
=> 3*e^-x - (3x +1)e^-x
=> 3*e^-x - f(x)
f''(x) = -3e^-x - f'(x)
=> -3e^-x - 3e^-x + f(x)
Now f(x) + 2f'(x) + f''(x)
=> f(x) + 2(3*e^-x - f(x)) + (-3e^-x - 3e^-x + f(x))
=> f(x) + 6e^-x - 2f(x) - 2*3e^-x + f(x)
cancel the common terms
=> 0
As f(x) + 2f'(x) + f''(x) = 0
=> f(x) = -2f'(x) - f''(x)
This proves that f(x) = -2f'(x) - f''(x)
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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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