Show that $1 - x^2 \le e^{-x^2}$ for $0\le x\le1$ and $e^{-x^2}\le1/(1+x^2)$ for $0\le x$ Found this question in my freshman year calc course textbook under elementary integration.
I used the following method:
$y=(1+t)e^{-t}$
$y'=-te^{-t}$
$y=1$ is maximum at $t=0$
taking $t=x^2$
$(1+x^2)e^{-x^2}\le1$
$e^{-x^2}\le1/(1+x^2)$
This is the second part of the question and I don't know how $x\ge0$ factors into this, shouldn't it work for all $x$? Any ideas on how I can prove the first part of the question?
This is what I have so far for the first part:
taking $t=-x^2$
$(1-x^2)e^{x^2}\le1$
$1-x^2 <= e^{-x^2}$
Again, don't know why it's only applicable for $0\le x\le1$
 A: I suspect that the strange restrictions on $x$ are meant to encourage you to make the substitution $t=x^2$ and reframe the two exercises as

*

*For $0\le t\le1$ show that $1-t\le e^{-t}$


*For $t\ge0$ show that $e^{-t}\le\frac{1}{1+t}$
Then if you know the expansions

*

*$e^{-t}=1-t+\dfrac{t^2}{2!}-\dfrac{t^3}{3!}+\cdots$


*$e^t=1+t+\dfrac{t^2}{2!}+\dfrac{t^3}{3!}+\cdots$
you will be able to verify the inequalities.
For the first inequality, if $t\in[0,1]$, then for $k\ge2$, it is the case that $\dfrac{t^k}{k!}>\dfrac{t^{k+1}}{(k+1)!}$
So $1-t\le1-t+\dfrac{t^2}{2!}-\dfrac{t^3}{3!}+\cdots=e^{-t}$ [equal when $t=0$]
The second inequality is equivalent to $1+t\ge e^t$ which follows immediately from the expansion for $e^t$.
A: Yes, it should work for all $x$ for the second part. Regarding the first part, let $f(x) = 1 - x - e^{-x}, x \ge 0$. Then $f'(x) = -1+e^{-x} = 0 \iff x = 0$. Also, $f''(0) = -1 < 0$. Thus $x = 0$ gives a relative and hence absolute maxima since it's the only one critical value of $f$. Thus $f(x) \le f(0) = 0 \implies 1 - x \le e^{-x}$. Since this is true for $x \ge 0$, it is true that $f(x^2) \le 0$. Thus $1-x^2-e^{-x^2} \le 0$. Thus $1-x^2 \le e^{-x^2}$.
