What is the f(θ)=2cos(θ)+cos^2(θ),0<= θ<= 2pi ?

The f(θ)=2cos(θ)+cos^2(θ),0<= θ<= 2pi is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

f(θ)=2cos(θ)+cos^2(θ),0<= θ<= 2pi

f (θ) = 2 cos (θ) + cos^{2} (θ), 0 \leq θ \leq 2 π

Period : 2 π

Domain : 0 \leq θ \leq 2 π

Range : - 1 \leq f (θ) \leq 3

X Intercepts : (\frac{π}{2}, 0), (\frac{3 π}{2}, 0), Y Intercepts : (0, 3)

Asymptotes : None

Extreme Points : Maximum (0, 3), Minimum (π, - 1), Maximum (2 π, 3)

Interval Notation

Domain : [0, 2 π]

Range : [- 1, 3]

Solution steps

Periodicity of

2 cos (θ) + cos^{2} (θ) : 2 π

Domain of

2 cos (θ) + cos^{2} (θ), 0 \leq θ \leq 2 π : 0 \leq θ \leq 2 π

Range of

2 cos (θ) + cos^{2} (θ), 0 \leq θ \leq 2 π : - 1 \leq f (θ) \leq 3

Axis interception points of

2 cos (θ) + cos^{2} (θ), 0 \leq θ \leq 2 π :

X Intercepts

: (\frac{π}{2}, 0), (\frac{3 π}{2}, 0),

Y Intercepts

: (0, 3)

Asymptotes of

2 cos (θ) + cos^{2} (θ), 0 \leq θ \leq 2 π :

None

Extreme Points of

2 cos (θ) + cos^{2} (θ), 0 \leq θ \leq 2 π :

Maximum

(0, 3),

Minimum

(π, - 1),

Maximum

(2 π, 3)

f (x) = (0.5)^{3 - x}

f (x) = (4 x^{3} + 1)^{2}

f (x) = \frac{x + 2}{2 - x - 1}

f (x) = - (x - 2) (x - 4)

g (x) = ∣ 2 x + 3 ∣

What is the f(θ)=2cos(θ)+cos^2(θ),0<= θ<= 2pi ?
The f(θ)=2cos(θ)+cos^2(θ),0<= θ<= 2pi is