Show that $f:\mathbb R\to\mathbb R,\ x\mapsto e^x-x^2-2x-2$ has exactly one zero.
First of all, using the IVT it is very easy to show that at least one zero must exist, i.e. in the interval $[2,100]$. However, I find it tricky to show that there can't be another zero. Using the taylor series of $e^x$ one can easily see that because $$f'(x)=-1-x+\frac{x^2}{2}+\frac{x^3}{6}+\dots>0,\quad \text{when } x>2,$$ the function is strictly monotone and therefor can't have another zero in $[2,\infty)$. But how can I show that it doesn't have a zero in $(-\infty,2)$? I can't seem to find a proper mathematical way to show this.
Best Answer
$f'(x) = 0$ occurs when $e^x = 2x + 2$. By plugging in $x = 0$, we see that the line $y = 2x + 2$ is not below the convex curve $y = e^x$, so $e^x = 2x + 2$ has two solutions, call them $x_1$ and $x_2$ where $x_1 < x_2$. These are the points where $f'(x) = 0$. $f(x)$ is increasing for $x < x_1$ and $x > x_2$, and is decreasing for $x_1 < x < x_2$.
For $x = x_1$ or $x_2$, we have $f(x) = e^x - x^2 - 2x - 2 < e^x - 2x - 2 = 0$. So $f(x_1)$ and $f(x_2)$ are negative. As a result, since $f(x)$ is increasing for $x < x_1$ and decreasing for $x_1 < x < x_2$, $f(x)$ cannot have any zeroes for $x < x_2$.
Since $f(x)$ is increasing on $x > x_2$, it can have at most one zero for $x > x_2$. Since $f(x_2) < 0$ and $f(x) > 0$ for large enough $x$, there is in fact one zero of $f(x)$ for $x > x_2$.
We conclude $f(x)$ has exactly one zero, occurring for some $x > x_2$.