Below is a graph of the function y = csc ( b x ) y=\csc\left(bx\right) y = csc ( b x ) . Determine the value of b .

Welcome back, everyone. So let's give this problem a try. Here we're told below is a graph of the function y is equal to the cosecant of b x, and we're asked to determine the value of b. Now to solve this problem, what I'm first going to do is draw a graph of something I'm a bit more familiar with. Now what I recognize here is that the cosecant is a reciprocal identity, and it's a reciprocal identity to the sine function. So I'm going to draw a graph of y is equal to the sine of bx. Now Now the way that I can figure out what this graph looks like is by recognizing that the sine function is going to be a wave that starts at the origin. And it's going to have its peaks where we have these curves that point upward these kind of smiley faces, and it's going to have the valleys where we have these curves that point downward these kind of frowny faces. So when we have our sine function go up here, it's gonna reach a peak, then it's gonna come down here in valley, then it's going to go through here, and then it's gonna reach another peak there, and then reach another valley right about there. So this is what the sine function looks like. Now what I can see for the sine function is we can start here at the origin, and we go up, and then down on this wave, and then back up, and complete a full period at 3 pi over 2 on the x axis. So what that tells me is that our period for this function is 3 pi over 2. Now the reason that this is helpful to find the period is because the period for a sine function is always going to be 2 pi over b. And since we just figured out that the period is 3 pi over 2, I can set 3pi over 2 equal to 2pi over b. And all I need to do from here is solve for b, so what I'm first going to do is multiply both sides of the equation by b. This is gonna get the b's to cancel on the right side, giving me 3pi over 2 times b is equal to 2pi. Now from here what I'm going to do is multiply both sides of the equation by 1 over pi. And this is going to just allow me to cancel this pi on both sides. Because notice that the pi's are going to cancel right there, and the pi's are going to cancel right here. So what we're going to have is 3 over 2 times b is equal to 2. The way that I'm going to solve this problem is I'm going to take both sides of this equation and multiply by the reciprocal of 3 over 2, which is 2 over 3. Now when I do this, this is going to allow me to cancel the threes right there, and the twos right here, leaving me with just 'b'. And then on the right side of this equation we're going to have 2 times 2, which is 4 divided by 3. So the solution for b is 4 thirds, and that's the answer to this problem. So hope you found this video helpful. Thanks for watching.

Below is a graph of the function y = csc ⁡ ( b x ) y=\csc\left(bx\right) y = csc ( b x ) . Determine the value of b .

b = 3 π 4 b=\frac{3\pi}{4} b = 4 3 π

0. Functions

Graphs of Trigonometric Functions

Calculus

0. Functions

Graphs of Trigonometric Functions

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Multiple Choice

Below is a graph of the function $y=\csc\left(bx\right)$ . Determine the value of b.

$b=\frac12$

$b=3$

$b=\frac43$

$b=\frac{3\pi}{4}$

Verified step by step guidance

The function given is y = csc(bx), which is the cosecant function. The cosecant function is the reciprocal of the sine function, so y = 1/sin(bx).

The period of the cosecant function is determined by the period of the sine function, which is 2π. For y = csc(bx), the period is given by 2π/b.

From the graph, observe that the function completes one full cycle between x = π/2 and x = 3π/2. This indicates that the period of the function is π.

Set the period 2π/b equal to the observed period π: 2π/b = π.

Solve for b by multiplying both sides by b and then dividing by π: b = 2.

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Struggling with Calculus?

Below is a graph of the function y=csc(bx)y=\csc\left(bx\right)y=csc(bx). Determine the value of b.

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Graphs of Trigonometric Functions practice set

Below is a graph of the function $y=\csc\left(bx\right)$ . Determine the value of b.