-/4 points details my notes 0/50 submissions used find all points on the graph of the function f(x)=2…

Explanation:

Step1: Find the derivative of $f(x)$

Using the sum - rule and chain - rule, if $y = 2\cos(x)+(\cos(x))^{2}$, then $y^\prime=f^\prime(x)=- 2\sin(x)+2\cos(x)(-\sin(x))=-2\sin(x)(1 + \cos(x))$.

Step2: Set the derivative equal to zero

Since the tangent line is horizontal when $f^\prime(x) = 0$, we set $-2\sin(x)(1+\cos(x)) = 0$. This gives two cases: Case 1: $\sin(x)=0$. In the domain $x\in[0,2\pi)$, when $\sin(x)=0$, $x = 0$ or $x=\pi$ or $x = 2\pi$. But since our domain is $[0,2\pi)$ (excluding $2\pi$), $x = 0$ or $x=\pi$. Case 2: $1+\cos(x)=0$, then $\cos(x)=-1$. In the domain $x\in[0,2\pi)$, when $\cos(x)=-1$, $x=\pi$.

Step3: Find the corresponding $y$ - values

When $x = 0$, $y=f(0)=2\cos(0)+(\cos(0))^{2}=2\times1 + 1^{2}=3$. When $x=\pi$, $y=f(\pi)=2\cos(\pi)+(\cos(\pi))^{2}=2\times(-1)+(-1)^{2}=-2 + 1=-1$.

Answer:

$(0,3)$ $(\pi,-1)$