Show that $f:\mathbb R\to\mathbb R$ has exactly one zero 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.
 A: $f'(x) = 0$ 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 $x > x_2$.
A: We have $$f(x)=e^x-x^2-2x-2$$ $$f'(x)=e^x-2x-2$$ $$f''(x)=e^x-2$$ $$f'''(x)=e^x$$
That means that $f'(x)$ has a global minimum at $x=\ln(2)$. One root of $f'(x)$ is between $-1$ and $0$ , the other one between $1$ and $2$
$f(x)$ is negative for $x=-1,0,1,2$
$f(x)$ is increasing outside the interval $[-1,2]$ and decreasing in the interval $[0,2]$. In the interval $(-1,0)$ , we have $|x^2+2x+2|\ge 1$, hence $f(x)$ is negative in this interval. $f(x)$ tends to $-\infty$ , if $x$ tends to $-\infty$. That means that $f(x)$ is negative for $x\le 2$. For $x\ge 2$, $f(x)$ is strictly increasing and tends to $\infty$ , if $x$ tends to $\infty$.
Hence $f(x)$ has exactly one real root.
A: Prove that f   is negative for $x<0$  then prove that it attains zero derivative at one and only one point for positive $x$  .  (The function at the point must be negative because f is monotonic for x greater than equal to 2 ) and since f(2) is negative we will agin reach a contradiction if the point is not negative due to Rolle’s THM) . Now if f had another root then we would again reach a contradiction due to Rolle’s thm since f is negative for x<0)
A: We have
$$f(x) = 0 \iff e^x-1 = (x+1)^2.$$
Clearly $0$ isn't a root. Notice that $(x+1)^2 \geq 0$ while $e^x - 1 < 0$ when $x< 0$ so there are no roots on $(-\infty,0].$
Notice that at any given root $x > 0$ we have
$$f'(x) = e^x - 2x - 2 = f(x) + x^2 = 0 + x^2 >0.$$
This means that $f$ (which is continuous) is strictly increasing on an open neighbourhood of each of it's roots so it can't have several roots.
