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So far all I have is this:

Let $f$ be a function where $f(x)=ex-e^x\leq 0$

$f'(x)=e-e^x \leq 0$, so $f$ is decreasing.

I'm stuck here. Can someone help me with the next steps?

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    $\begingroup$ You've already made one mistake: $e - e^x$ is not always less than 0. $\endgroup$ – Peter Shor Apr 22 '15 at 4:36
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Let $f(x) := e^{x} - ex$ for all $x \in \mathbb{R}.$ We claim that $f \geq 0$ on $\mathbb{R}.$

Note that $f(1) = 0,$ that $f'(x) > 0$ for all $x > 1$, and that $f'(x) < 0$ for all $x < 1.$ Thus $f$ is increasing on $]1, +\infty[$ so that $f > 0$ on $]1, +\infty[.$ Since $f$ is decreasing on $]-\infty, 1[$ and is continuous on $\mathbb{R},$ it follows that $f > 0$ on $]-\infty, 1[.$

Including the case where $f(1) = 0,$ we thus have $f \geq 0$ on $\mathbb{R}.$

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Hint: $e^x$ is convex, hence stays above its tangent at $x=1$.

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    $\begingroup$ people down voting any answer should explain, i found this as an amazing hint $\endgroup$ – avz2611 Apr 22 '15 at 5:02
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    $\begingroup$ Though I am not the person down voted you, I doubt that the OP knows the connection with convexity... $\endgroup$ – Megadeth Apr 22 '15 at 5:07
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First of all, $f'(x)$ isn't always less than $0$. It's less than/equal to $0$ at $x=1$ however. At $x=1$, $f(x)=0$, and if it's decreasing it will always be less than $0$. For $x<1$, it shouldn't be hard to show.

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There are $3$ cases to consider:

$a):$ $x < 0 \Rightarrow ex < 0 < e^x \Rightarrow ex < e^x$.

$b):$ $ 0 \leq x \leq 1$, $f(x) = ex - e^x \to f'(x) = e - e^x \geq 0 \to f(x) \leq f(1) = e - e = 0 \Rightarrow ex - e^x \leq 0 \Rightarrow ex \leq e^x$.

$c): $ $ 1 < x $: $f(x) = ex - e^x \to f'(x) = e - e^x < 0 \to f(x) < f(1) = e - e = 0 \to ex < e^x$.

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