The function to be graphed is: y=−2+4cos(12x−π3) y = -2 + 4 \cos \left(\frac{1}{2} x - \frac{\pi}{3}\right) y=−2+4cos(21x−3π)
The cosine function completes one cycle over an interval of 2π2\pi2π. Given the argument of the cosine function is 12x−π3\frac{1}{2} x - \frac{\pi}{3}21x−3π, we set: 12x−π3=0to12x−π3=2π \frac{1}{2} x - \frac{\pi}{3} = 0 \quad \text{to} \quad \frac{1}{2} x - \frac{\pi}{3} = 2\pi 21x−3π=0to21x−3π=2π Solving for xxx: 12x−π3=0 ⟹ x=2π3 \frac{1}{2} x - \frac{\pi}{3} = 0 \implies x = \frac{2\pi}{3} 21x−3π=0⟹x=32π 12x−π3=2π ⟹ 12x=2π+π3 ⟹ x=4π+2π3=14π3 \frac{1}{2} x - \frac{\pi}{3} = 2\pi \implies \frac{1}{2} x = 2\pi + \frac{\pi}{3} \implies x = 4\pi + \frac{2\pi}{3} = \frac{14\pi}{3} 21x−3π=2π⟹21x=2π+3π⟹x=4π+32π=314π Thus, the range for one complete cycle is: x∈[2π3,14π3] x \in \left[\frac{2\pi}{3}, \frac{14\pi}{3}\right] x∈[32π,314π]
The cosine function ranges from -1 to 1. Therefore: y=−2+4cos(12x−π3) y = -2 + 4 \cos \left(\frac{1}{2} x - \frac{\pi}{3}\right) y=−2+4cos(21x−3π) ymin=−2+4(−1)=−6 y_{\text{min}} = -2 + 4(-1) = -6 ymin=−2+4(−1)=−6 ymax=−2+4(1)=2 y_{\text{max}} = -2 + 4(1) = 2 ymax=−2+4(1)=2 Thus, the y-axis range is: y∈[−6,2] y \in [-6, 2] y∈[−6,2]
{"axisType": 3, "coordSystem": {"xmin": 2.0944, "xmax": 14.6608, "ymin": -6, "ymax": 2}, "commands": ["y = -2 + 4_cos((1/2)_x - 3.1416/3)"], "latex_expressions": ["y=−2+4cosleft(frac12x−fracpi3right)y = -2 + 4 \\cos \\left(\\frac{1}{2} x - \\frac{\\pi}{3}\\right)y=−2+4cosleft(frac12x−fracpi3right)"]}
Oops, Image-based questions are not yet availableUse Solvely.ai for full features.
Failed. You've reached the daily limit for free usage.Please come back tomorrow or visit Solvely.ai for additional homework help.