Find range of composite trigonometric function Find the range of the function:
$$f(x)= \cos^2x-\cos x $$
Answer is: $ [-1/4 ,2]$
I've factorized $\cos x$ and thus got min and max values equal to $[0,2]$ using inequalities. I know it's $-1/4$ because of $\cos(\pi/3).$ Any help or tip is appreciated, thanks.
 A: $$f(x)=\cos^2 x-\cos x=\left(\cos x -\frac{1}{2}\right)^2-\frac{1}{4}\ge -\frac{1}{4}$$
with equality if and only if $\cos x=\frac{1}{2}$, i.e. $x=\pm\frac{\pi}{3}+2\pi k$ for some $k\in\mathbb Z$.
Also remember $\cos x\in[-1,1]$ for all $x\in\mathbb R$ by the definition of $\cos x$,
so $\cos^2 x-\cos x\le 2$ with equality if and only if $\cos x=-1$, i.e. $x=\pi+2\pi k$ for some $k\in\mathbb Z$.
$f(x)$ is also continuous in $\mathbb R$, so its range is $\left[-\frac{1}{4},2\right]$.
A: because $\cos x \in [-1,1]$ for any $x \in \mathbb R $ let be 
$ \displaystyle f:[-1,1] \rightarrow \mathbb R, f(t)=t^2-t$ 
this quadratic function have a minimum in $t_V=\frac{-b}{2a}=\frac{1}{2}$ and the minimum is $f(t_V)=f(\frac{1}{2})=\frac{-1}{4}$
the maximum is the bigger value between $f(-1)=2$ and $f(1)=0$
so the range of $f(x)$ is $[\frac{-1}{4},2]$

A: If you derivate to find the extrems you get $f'(x) = (1-2\cos(x))\sin(x)$, so $\cos(x_1)=1/2$ or $\sin(x)=0$ (which means $\cos(x_2)=1$ or $\cos(x_3)=-1$). (indexing just for reference)
Testing for the 3 extreams you get:
$$f(x_1) = - 1/4$$ $$f(x_2)=0$$ $$f(x_3)=2$$
so it span [-1/4;2] with max and min over the x_1 and x_3 values (which are 'more than one' each).
A: $f'(x) =-2cos(x)sin(x)+sin(x)$
$-2cos(x)sin(x)+sin(x) = 0$
$sin(x)(1-2cos(x))=0$
$sin(x) = 0$ or $cos(x)=\dfrac{1}{2}$
$x = 0 + k\pi$ or $cos(x_0)=\dfrac{1}{2}$
$f(0) = 0 ; f(\pi) = 2 ; f(x_0)= \dfrac{1}{4}-\dfrac{1}{2} = \dfrac {-1}{4}$
