Inverse function to f

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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 $0$.