<s>[INST] 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.
 [/INST] $\sin(x) + \cos^2(x) = - \sin^2(x) + \sin(x) + 1$ 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 $x = \frac{\pi}{6}$.
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