How can I argue that $f(x) = (2-x)^3-x+\frac{3}{2}$ is decreasing with a polynomial? 
$$f(x) = (2-x)^3-x+\frac{3}{2}$$

I have to give a mathematical argument for the said function in the title being decreasing.
My first thought was finding $f'(x)$ but I got a polynomial and for that reason I don't know how to argue that it is decreasing since $f'(x)<0$ has to be true. What I mean is that I can't see it in the polynomial, which is:
$f'(x)=-3x^2+12x-13$.
However, the second derivative gives me something I can work with: $f''(x)=-6x+12$.
So my question is, how do I argue that $f(x) = (2-x)^3-x+3/2$ is decreasing with the first derivative being a polynomial, $f'(x)=-3x^2+12x-13$?
 A: $$f'(x)=-3x^2+12x-13 = -3(x^2-4x+4)-1 = -3(x-2)^2-1<0 \text{ for all } x\in\mathbb R$$
A: $f’(x)=-3x^2+12x-13$ tells us that we need to show that a quadratic function is always negative. Thus we must analyse its discriminant, which is ($\Delta=b^2-4ac$ for $ax^2+bx+c$).
Here we get $\Delta=12^2-4(-13)(-3)=144-4\times 39=-12$. But when discriminant is negative, there are no real roots, so the graph cannot cut (not even touch) the X-axis. Thus the polynomial $f’(x)=-3x^2+12x-13$ must always be either positive or negative. You can determine this in two ways:

*

*Check a simple value, like $f’(0)$, which here  is $-13<0$. Thus the polynomial $f’(x)=-3x^2+12x-13$ must always be negative.

*If a polynomial $ax^2+bx+c$ has discriminant <$0$ then if the leading coefficient $a$ is positive, then the polynomial is always positive, and vice versa, which happens here ($-3<0$).


If we were to continue with your method, $f’’(x)=-6x+12$. $f’’(x)$ is positive in $(-\infty, 2)$ and negative in $(2,\infty)$. This implies  that $f’(x)$ increases in $(-\infty, 2)$ and decreases in $(2,\infty)$. So the function $f’(x)=-3x^2+12x-13$ has $\displaystyle\lim_{x\to -\infty}f(x)=-\infty$ and increases with $x$ up till the value $f’(2)$, from where it starts decreasing, again to $\displaystyle\lim_{x\to +\infty}f(x)=-\infty$.
But the value of $f’(2)$ is $-1$. This means that the values of $f’(x)$ always lie in $(-\infty,0)$ so $f’(x)$ is always negative.
