Range of $\sin^2x-\sin x +1$ How can we find the range of $f(x) =\sin^2x-\sin x +1$? 
The function can be written as $(\sin x-\frac{1}{2})^2+\frac{3}{4}$.
Range of $\sin x$ function is given by : $-1 \leq \sin x \leq 1$.  Please guide how to get the result.

So, $-3/2 \le \sin(x) -1/2 \le 1/2$. Squaring, $0\le (\sin(x)-1/2)^2 \le 9/4$.  Now add $3/4$ to see that $3/4 \le (\sin(x) - 1/2)^2 + 3/4 \le 3$.  So we know that the range lies in the interval $[3/4, 3]$.  Now $sin(3\pi/2) = -1$ so
$f(-3\pi/2) = 9/4$.  We have $\sin(\pi/3) = 1/2$ so $f(\pi/3) = 3/4$.  We now know that $3$ and $3/4$ lie in the range of $f$.  By the intermediate value theorem, the range of $f$ is the whole interval.
