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Question
Show that f(x) = (x − 1) ex + 1 is an increasing function for all x > 0 ?
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Solution
\[f\left( x \right) = \left( x - 1 \right) e^x + 1\]
\[f'\left( x \right) = \left( x - 1 \right) e^x + e^x \]
\[ = x e^x - e^x + e^x \]
\[ = x e^x \]
\[\text { Given }:x > 0 \]
\[\text { We know,}\]
\[ e^x > 0\]
\[\Rightarrow x e^x > 0\]
\[ \Rightarrow f'\left( x \right) > 0, \forall x > 0\]
\[\text { So },f(x)\text { is increasing on for all }x>0.\]
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