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Q.
If the function $f ( x )= cx \cdot e ^{- x }-\frac{ x ^2}{2}+ x$ is decreasing for every $-\infty< x \leq 0$, then the least value of $c ^2$ is equal to
Application of Derivatives
Solution:
$f ( x )= c x e ^{- x }-\frac{ x ^2}{2}+ x$, we must have, $f ^{\prime}( x ) \leq 0 \forall x \leq 0$
$\Rightarrow (1- x )\left( c e ^{- x }+1\right) \leq 0 \forall x \leq 0 \Rightarrow c \leq- e ^{ x } \forall x \leq 0$
so, $c \in(-\infty,-1] \Rightarrow $ Least value of $c ^2=1$