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I have a function like this: $$f(x)=\frac{1-e^{-bx}}{x},$$ where $b>1$.

Is this function a monotonic function?

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try plotting this in wolfram and you will see*x%29%29%2Fx). Can you tell now? – jay-sun Nov 28 '12 at 16:58
Yes, I have plotted, too. But how to prove it? – windy Nov 28 '12 at 17:03
up vote 2 down vote accepted

Observe that $$f'(x)=\frac{bxe^{-bx}-\left(1-e^{-bx}\right)}{x^2}=\frac{(bx+1)e^{-bx}-1}{x^2}.$$

Now, since $x=0$ isn't in our domain, then $f'(x)=0$ if and only if $(bx+1)e^{-bx}=1$ if and only if $bx+1=e^{bx}$. Note that $y=bx+1$ is the tangent line to $y=e^{bx}$ at $x=0$, so since $y=e^{bx}$ is a convex function, then these two curves never again intersect, and in particular $bx+1<e^{bx}$ for all non-zero real $x$. From this, what can we conclude about the sign of $f'(x)$ for all non-zero real $x$?

As you pointed out in the comments below, we never used the assumption that $b>1$ above, so we can do even better! The argument above actually works for any real $b$, not just $b>1$, nor even just $b>0$.

In fact, given $b\neq 0$, if we define $$g(x)=\begin{cases}b & x=0\\f(x) & \text{otherwise,}\end{cases}$$ we get a continuously differentiable extension of $f(x)$ defined on all of $\Bbb R$. (Obviously, we can't do that when $b=0$, for then $f(x)=\frac1x$, which can't be fixed to a continuous function on $\Bbb R$, much less a continuously differentiable one.) In particular, we have $$g'(x)=\begin{cases}-\frac{b^2}2 & x=0\\f'(x) & \text{otherwise,}\end{cases}$$ so we can see that $g(x)$ is monotone decreasing on all of $\Bbb R$.

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Thank you, so I can have a conclusion that for $x>0$ and $b>0$, f(x) is a monotonic decreasing function. – windy Nov 28 '12 at 17:30
In your answer, I find that $b>0$ also holds the conclusion, am I right? – windy Nov 28 '12 at 17:55
You're quite right! I'll go ahead and expand my answer accordingly. – Cameron Buie Nov 28 '12 at 18:07
OK, Thanks for your answer. – windy Nov 28 '12 at 18:16

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