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Values of parameter

Find all values of \lambda \in \mathbb{R} such that

    \[e^x - \lambda x \geq 0 \quad \text{forall} \; x \in \mathbb{R}\]

Solution

We’re invoking the same technique as in the problem hereSuccessively we have:

    \begin{align*} e^x - \lambda x \geq 0 &\Leftrightarrow e^x \geq \lambda x \\ &\Leftrightarrow \lambda x e^{-x} \leq 1 \\ &\Leftrightarrow f(x) \leq 1 \end{align*}

Obviously f is differentiable in \mathbb{R} and its derivative is

    \[f'(x) = \lambda e^{-x} \left ( 1-x \right )\]

It is f'(x) =0 \Leftrightarrow x=1. We distinguish cases:

  • If \lambda>0 then f attains global maximum at x=1 equal to f(1)=\frac{\lambda}{e}. Since f(x) \leq 1 it follows that \lambda \leq e.
  • If \lambda<0 then f attains global minimum at x=1 equal to f(1)=\frac{\lambda}{e}. In this case , however , the inequality cannot hold for all \lambda <0; since f is continuous its range is:

        \begin{align*} f(\mathbb{R}) &= \left [ f(1), \lim_{x \rightarrow -\infty} f(x) \right ) \cup \left [ f(1), \lim_{x \rightarrow +\infty} f(x) \right ) \\ &= \left [ \frac{\lambda}{e}, +\infty \right ) \cup \left [ \frac{\lambda}{e} , 0 \right ) \\ &= \left [ \frac{\lambda}{e}, +\infty \right ) \end{align*}

    Hence this case is rejected.

  • For \lambda=0 the inequality obviously holds for all x \in \mathbb{R}.

Summing up , \lambda \in [0, e].

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