Math Help - range

1. range

Find Range of $f(x) = \sin^2 x-5\sin x+6$

where $x\in \left[0,\pi\right]$

2. Re: range

Originally Posted by jacks
Find Range of $f(x) = \sin^2 x-5\sin x+6$

where $x\in \left[0,\pi\right]$
$f'(x)=2\sin x \cdot \cos x-5 \cos x=\cos x \cdot (2\sin x-5)$

In order to find minimum of function we have to solve following equation :

$f'(x)=0 \Rightarrow \cos x=0 \Rightarrow x= \frac{\pi}{2} \Rightarrow \sin x = 1$

So ,we can conclude that : $f_{min}=2$

On the other hand function has maximum for $x=0$ , and $x=\pi$ , therefore : $f_{max}=6$

So for $x \in [0,\pi]$ we have : $f(x) \in [2,6]$