Let $ f ∈ C[a, b]$ be differentiable in $(a, b)$. If $f(a) = f(b) = 0$, then, for any real number $α$, there exists $x ∈(a, b)$ such that $f'(x) + αf(x) = 0$.
I think Rolle's theorem is of some use here, but am not sure. Any help. Thanks beforehand.
Is the statement about differentiable function true?
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calculus
real-analysis
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3Multiplying through by the integrating factor $e^{\alpha x}$, we find that it is enough to find $x$ such that the derivative of $e^{\alpha x} f(x)$ vanishes. But Rolle's theorem implies this. QED. – 2017-01-01
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0If you let $g(x)=e^{\alpha x}f(x)$ then $g(a)=e^{\alpha a}f(a)=0$. Similarly $g(b)=0$. So by Rolle's theorem there exists $x \in (a,b)$ such that $g'(x)=0$. – 2017-01-01
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Of course this directly follows from Rolle's theorem. Note that by solving the equation, you get $f(x)=C e^{-\alpha x}$ for some arbitrary constant $C.$