Applications of the derivative. Strictly monotonic functions How I can prove this?
Let $$f(x) = \sin x + \frac{x^2}{2} - 2x.$$ 
Show that $f$ has only one critical number in the interval $[2, 3]$.
It belongs to the chapter on applications of the derivative. Section: Relative extrema of real functions.
 A: Hint: The second derivative is positive on $[2,3]$.   To see why, notice $$f''(x) =-\sin x+1>0$$ and that  $|\sin x |<1$ on $[2,3]$.
This means that the derivative $f^'(x)$ is increasing on $[2,3]$, and you can use this idea to find that there is at most one critical number.  The intermediate value theorem tells you there is exactly one.
A: The definition of "critical point" is that it is a point where the function is defined, but the derivative is either equal to $0$ or undefined. 
So, let's try taking the derivative and staring at it, perhaps we can "tell" immediately that there is one and only one critical point in $[2,3]$.
$$\begin{align*}
f(x) &= \sin x + \frac{x^2}{2} -2x\\
f'(x) &= \cos x + x - 2.
\end{align*}$$
This is defined everywhere, so the only critical points will be stationary points.
So... can we just "tell" that $f'(x)$ takes the value $0$ only once on $[2,3]$ by staring at it?  Hmmm... That doesn't seem easy to do (you could note that $x-2$ takes the values from $0$ to $1$, and that $\cos x$ is negative and decreasing, but that tells you that $f'(x)$ is negative at $2$ and positive at $3$, hence $f$ has at least one critical point by the Intermediate Value Theorem since $f'$ is continuous, but it doesn't tell you it has exactly one).
What else can we do? How about seeing whether this function is strictly monotone? As noted parenthetically, $f'(2)\lt 0$ and $f'(3)\gt 0$. If $f'(x)$ is strictly monotone on $[2,3]$, then it goes through zero exactly once. 
