# Inverse function to f

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.

• 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. – gd1035 Oct 5 '18 at 17:55
• it does have an inverse but not everywhere – Vasya Oct 5 '18 at 17:58

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 $$I=[-\ln 2,\infty).$$

• Thanks, does is mean that the function is strictly Increasing in interval (−ln2,∞), since interval has ∞? – Fork Oct 5 '18 at 20:22
• 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 $\infty.$ – user376343 Oct 5 '18 at 20:26
• 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. – Fork Oct 5 '18 at 21:08
• @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 $f.$ – user376343 Oct 5 '18 at 21:27

$$f'(x)=3e^{-3x}(2e^x-1)$$, 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.

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 $$0$$.