Write an arithmetic sequence that gives the nth positive x-intercept of the graph of f(x)=cos1/2x. Leave your answer in terms of π. WILL GIVE BRAINLIEST!

Accepted Solution

Let us consider the function given. It is [tex]y=cos \frac{1}{2}x [/tex]. The coefficient of x (here 1/2) gives the frequency of the curve. What this means is that we see 1/2 a cycle between x=0 and x=[tex]2 \pi [/tex]. Put another way, to see a full cycle we would need [tex] \frac{2 \pi }{ \frac{1}{2} }=4 \pi [/tex]. The period of the function given is [tex]4 \pi [/tex].

Let us think about a cycle of cosine. It starts (x=0) and ends ([tex]x=4 \pi [/tex]) its cycle at its highest value (here f(x)=1). It is at it's lowest (here f(x)=-1) in the middle ([tex]x=2 \pi [/tex]) it is between the highest and the lowest that it crosses the x-axis.

That is, it has an x-intercept between 0 and [tex]2 \pi [/tex] That is, at [tex] \pi [/tex]

The next one comes between [tex]2 \pi [/tex] and [tex]4 \pi [/tex]. That is, at [tex]3 \pi [/tex]

The attached shows the function and it's x-intercepts. You can see that they occur at: [tex] \pi ,3 \pi ,5 \pi ,7 \pi ,9 \pi ,...[/tex]. This is the arithmetic sequence that contains the x-intercepts of the function.

As you can see the nth zero will occur at [tex]n \pi [/tex]