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I have function f : $\mathbb{R}$$\mathbb{R}$ defined by f(x) = $e^{-3x}-3e^{-2x}$ and have found that f'(x)=$-3e^{-3x}+6e^{-2x}$. Can someone explain why f does not have an inverse function. And how can I find the largest interval I that containing the origin such that the function g: I → $\mathbb{R}$ given by g(x)= $e^{-3x}-3e^{-2x}$ has an inverse function.

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  • $\begingroup$ If you look at the graph of $f(x)$ you will see that it is not injective (1-1). So it doesn't have an inverse on all real numbers. $\endgroup$ – gd1035 Oct 5 '18 at 17:55
  • $\begingroup$ it does have an inverse but not everywhere $\endgroup$ – Vasya Oct 5 '18 at 17:58
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Recall that $e^t>0$ for $t \in \mathbb{R}.$

Thus the sign of $f'(x)=3e^{-3x}(2e^x-1)$ is that of $(2e^x-1),$ and we found that $f$ is

  • decreasing on $(-\infty,-\ln 2)$
  • increasing on $(-\ln 2,\infty)$

Note: We can include $(-\ln 2)$ to one of these intervals.

Therefore $f$ is not injective and doesn't have an inverse.

As $-\ln 2<0,$ the largest interval $I$ containing $0$ and satisfying that $f/I$ has an inverse function, is $I=[-\ln 2,\infty).$

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  • $\begingroup$ Thanks, does is mean that the function is strictly Increasing in interval (−ln2,∞), since interval has ∞? $\endgroup$ – Fork Oct 5 '18 at 20:22
  • $\begingroup$ The function is strictly increasing in $[-\ln 2,\infty)$ therefore also in $(-\ln 2, \infty).$ But I do not understand what should be relation to $\infty.$ $\endgroup$ – user376343 Oct 5 '18 at 20:26
  • $\begingroup$ I was thinking since strictly increasing function on interval (a,b) have f(b)>f(a), then function in this interval with ∞, should be strictly increasing. $\endgroup$ – Fork Oct 5 '18 at 21:08
  • $\begingroup$ @Fork revise your definition. "Strictly increasing" concerns ANY two numbers $x,y$ from $(a,b),$ and the values $f(x), f(y).$ Moreover, if the interval is open, it is not possible to upload extremities into $f.$ $\endgroup$ – user376343 Oct 5 '18 at 21:27
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$f'(x)=3e^{-3x}(2e^x-1)$, let $g(x)=2e^x-1$, $g'(x)=2e^x>0$ implies that $g$ is an increasing function. $g(x)=0$ is equivalent to $x=-ln(2)$.

We deduce that if $x\leq -ln(2)$ $f$ decreases, if $x\geq -ln(2)$ $f$ increases. $lim_{x\rightarrow -\infty}f(x)=+\infty$, $lim_{x\rightarrow +\infty}f(x)=0$, $f(-ln(2))<0$. Let $c>0$ such that $c<|f(-ln(2))|$, IVT implies that there exists $x<-ln(2), y>-ln(2)$ such that $f(x)=f(y)$ so $f$ is not injective.

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Note that a function can have an inverse on a certain domain iff it is injective on that region. Note that here. $f$ is not injective on all of $\mathbb{R}$; in fact, in this case, you'll notice that there's a single minima an on either side, the function is injective. So if the minima is $x_0$, your required interval is essentially $[x_0,\infty)$ or $(-\infty,x_0]$. To find which of these is the answer you want, check what $x_0$ actually is and choose the appropriate interval that contains $0$.

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