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.
 A: As $\sin x$ varies through $[-1,1]$, $\sin x-\frac12$ varies through $[-\frac32,\frac12]$, so $(\sin x-\frac12)^2$ varies through $[0,\frac94]$, so $(\sin x-\frac12)^2+\frac34$ varies through $[\frac34,3]$.
A: 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.
A: $-1 \leq sinx \leq 1 \Rightarrow -\frac{3}{2} \leq sinx-\frac{1}{2} \leq \frac{1}{2} \Rightarrow 0 \leq (sinx - \frac{1}{2})^2 \leq \frac{9}{4} \Rightarrow \frac{3}{4}\leq(sinx-\frac{1}{2})^2 + \frac{3}{4}\leq 3$
A: The function $\ (\sin x-\frac{1}{2})^2+\frac{3}{4}\ $ is continous and bounded. 
Let's take a look at the extrema. These lie at the points where the derivative $\ 2\cos x\cdot(\sin x-\frac{1}{2})\ $ vanishes. So
$\sin x-\frac{1}{2}=0\ \ \ \ \ \ \text{or}\ \ \ \ \ \ \cos x=0\ \rightarrow\ \sin x=\pm1.$
Hence the extrema take the values $\ 0^2+\frac{3}{4}=\frac{3}{4}\ $ and $\ ((-1) -\frac{1}{2})^2+\frac{3}{4}=3$.
