Graph f(x)=1/(1-2cos(x))

Problem

ƒ(x)=1/(1−2*cos(x))

Solution

1. Identify the domain by finding where the denominator is zero.

1−2*cos(x)=0

cos(x)=1/2

x=π/3+2*n*π

x=−π/3+2*n*π

The function has vertical asymptotes at these values.

2. Determine the period of the function. Since the only trigonometric term is cos(x) the period is 2*π

3. Find the y-intercept by evaluating the function at x=0

ƒ(0)=1/(1−2*cos(0))

ƒ(0)=1/(1−2*(1))

ƒ(0)=−1

4. Find the local extrema by analyzing the range of the denominator. The expression 1−2*cos(x) oscillates between 1 - 2(1) = -1a*n*d - 2(-1) = 3.W*h*e*ncos(x) = 1,(x) = -1(l*o*c*a*l*m*a*x*i*m*u*m).W*h*e*ncos(x) = -1,(x) = \frac{1}{3}$ (local minimum).

5. Analyze the behavior near the asymptotes. As x approaches π/3 from the left, cos(x) is slightly greater than 0.5 making the denominator negative and ƒ(x) approach −∞ As x approaches π/3 from the right, cos(x) is slightly less than 0.5 making the denominator positive and ƒ(x) approach +∞

6. Sketch the graph using the calculated points and asymptotes. The graph consists of U-shaped and n-shaped curves separated by vertical asymptotes at x=±π/3,(5*π)/3,…

Final Answer

ƒ(x)=1/(1−2*cos(x))

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