graph f(x)=cos1/2x on the interval (-2pi,2pi)

Answer 1

Hello!

First, given the interval you are asking for it over, I'm going to assume you mean cos[(1/2)*x] = cos(x/2) as opposed to cos(1/2x).

Next, keep in mind the values where cos(θ) = 0; θ = π/2 ± nπ; where cos(θ) = 1; θ = ±2nπ; and where cos(θ) = -1; θ = π ± 2nπ. For each answer, n is any whole number.

Now we can compare f(x) and the portion inside of the cosine to the basic cosine function to find out where f(x) has its zeros [cos(x/2) = 0], highest points [cos(x/2) = 1], and lowest points [cos(x/2) = -1]. We can do this by setting θ = x/2 for each situation and solving for x.

Let's start with where f(x) has its zeros. Given that for cos(θ) = 0 we got θ = π/2 ± nπ...

x/2 = π/2 ± nπ

x = π ± 2nπ

Plugging in whole numbers for n and carrying out both addition and subtraction, we get the answers {-π, π} and therefore can plot the points (-π, 0) and (π, 0).

Let's do that again for f(x) at its maximum [f(x) = 1]. Given that for cos(θ) = 1 we got θ = ±2nπ...

x/2 = ±2nπ

x = ±4nπ

Plugging in whole numbers for n and considering both positive and negative values, we find only the answer {0} and therefore can plot the point (0,1).

Lastly, let's do that one more time for f(x) at its minimum [f(x) = -1]. Given that for cos(θ) = -1 we got θ = π ± 2nπ...

x/2 = π ± 2nπ

x = 2π ± 4nπ

Plugging in whole numbers for n and carrying out both addition and subtraction, we get the answers {-2π, 2π} and therefore can plot the point (-2π,-1) and (2π,-1).

With all our points plotted, we just need to connect the dots using the classic wave pattern for cosine to get our final answer.

I hope this helps!