Find range of $\sin x+\cos^{2}x$ By differentiating and equating to $0$ I know that the maximum must be $5/4$.   The minimum is where I am confused. $-1$ would be logical but I'm not sure if this function can ever be equal to $0$. Answer is apparently $1$ to $5/4$ but I think that is wrong.
 A: $$\sin x+\cos^2x=\sin x+1-\sin^2x=\frac{5-(2\sin x-1)^2}4$$
Now for real $\sin x,(2\sin x-1)^2\ge0$
and $-1\le\sin x\le1\implies-2-1\le2\sin x-1\le2-1$
$\implies(2\sin x-1)^2\le$max$((-2-1)^2,(2-1)^2)=9$
A: Note that our function has $2\pi$ as a period. So to identify the maximum and minimum values, we can confine attention to the interval $[0,2\pi]$. And because our function is continuous on this closed interval, it attains a maximum and minimum on this interval.
Differentiate and set the derivative equal to $0$. We get the two solutions $\cos x=0$ and $\sin x=\frac{1}{2}$. For the sake of caution, we should also consider the two endpoints $0$ and $2\pi$.
Evaluate our function at all the candidates. The maximum value is reached when $\sin x=\frac{1}{2}$, giving value $\frac{5}{4}$ and the minimum value is reached at $x=\frac{3\pi}{2}$, giving value $-1$. 
Since our function is continuous, by the Intermediate Value Theorem every value between the minimum and maximum is attained. 
A: Let us derive: $f'(x)=\cos x-2\cos x\sin x=\cos x(1-2\sin x)$.
The roots are $\color{red}{\cos x=0}$, hence $\color{red}{\sin x=\pm1}$ and $\color{green}{\sin x=1/2}$, hence $\color{green}{\cos x=\pm\sqrt3/2}$.
Let us evaluate $f(x)=\sin x+\cos^2x$ on these values. We get : $\color{red}{\pm1+0}$, and $\color{green}{1/2+3/4}$.
So the range is $[-1,5/4]$.

A: $\sin(x) + \cos^2(x) = - \sin^2(x) + \sin(x) + 1$ since $\cos^2(x) = 1 - \sin^2(x)$. Let $t=\sin(x)$. 
$\sin(x) + \cos^2(x) = -t^2 + t + 1$ and $t \in [-1,1]$ because $\sin(x)\in [-1,1]$. Draw the curve of the quadratic $-t^2 + t + 1$ for $t \in [-1,1]$. The quadratic $-t^2 + t + 1$ has a root $\frac{1-\sqrt{5}}{2}$ between $-1$ and $1$. The function $\sin(x) + \cos^2(x)$ therefore has a root at $\sin(x) = \frac{1-\sqrt{5}}{2}$ i.e., the function $\sin(x) + \cos^2(x)$ attains the value $0$ for $\sin(x)=\frac{1-\sqrt{5}}{2}$. The quadratic attains its minimum at $t = -1$ and has the value of $-1$. Hence, the original function too has the minimum value of $-1$ when $t = \sin(x) = -1$ or $x = -\frac{\pi}{2}$. The quadratic attains a maximum at $t=\frac{1}{2}$ and has the value $\frac{5}{4}$. The original function thus has a maximum value of $\frac{5}{4}$ at $t=\sin(x)=\frac{1}{2}$ or $x = \frac{\pi}{6}$.
A: $sin(x)+cos^{2}(x)=sin(x)+1-sin^{2}(x)$, so range is $f(\left\langle -1,1\right\rangle )$, where $f(x)=1+x-x^{2}$, i.e. Range $=<-1,5/4>$ 
