# Find lower bound of function

Can someone help me finding a lower bound to the function $$f(x)=\frac{x-1}{e^{-1}-xe^{-x^2}},$$ where $x\in[1,+\infty[$?

Taking the derivative and then solve $f'(x)=0$ isn't analytically possible. Then I tried the second best thing, find a lower bound but I don't really know how to start, so any help would be most welcome.

• Are you seeking to find a tight lower bound? – user40314 Sep 26 '13 at 15:27
• @user40314 In the sense that the bound should be close to the actual minimum?Well, yes. – PML Sep 26 '13 at 15:30
• Alpha finds $\approx 2.33815$ at $x \approx 1.34929$ – Ross Millikan Sep 26 '13 at 17:44
• Since $$-\frac{1}{f(x)}=\frac{xe^{-x^2}-e^{-1}}{x-1},$$ we have that if $f$ obtains its lowest value at $x=a$ then the line connecting $(1,e^{-1})$ and $(a,ae^{-a^2})$ is tangent to the graph of $xe^{-x^2}$ at $a$. From here we get $$1-2a^2+2a^3=e^{a^2-1}.$$ That $\approx 1.34929$ is the only root of this equation with $a > 1$. – njguliyev Sep 26 '13 at 20:40
• @njguliyev Your last result of $f(\approx 1.34929)$ being the minimum is taken from where? Numerical methods? I was hoping to find a lower bound from analysis. – PML Sep 26 '13 at 21:17

## 1 Answer

Following njguliyev's comment, consider
$$\frac{1}{f(x)}=\frac{g(x)-g(1) }{x-1}\tag{1}$$ where $g(x)=-xe^{-x^2}$. By the Mean value theorem, every value of $1/f$ on $(1,\infty)$ is also attained by $g'$ on $(1,\infty)$. So we need an upper bound on $g'$. Since $$g'(x) = (2x^2-1)e^{-x^2}$$ $$g''(x) = 2x(3-2x^2)e^{-x^2}$$ it follows that $g'$ is maximal when $x^2=3/2$. Thus, the maximum of $g'$ on $(1,\infty)$ is $2e^{-3/2}$. This gives an upper bound on (1), and consequently $$f(x) \ge \frac{e^{3/2}}2 ,\quad x\ge 1$$ This lower bound is $\approx 2.24$, not far from the minimum of $\approx 2.34$ found in comments numerically.