Question

What are the critical values (critical points) for one period starting at the horizontal shift of the following function?y = 2 cos(3x) - 4

Answer 1

We have the following function:

`y=2cos(3x)-4`

And we have to find the critical values (critical points) for one period starting at the horizontal shift of the given function.

1. We can see that the general function for a horizontal shift in the cosine function is given by:

`\begin{gathered} y=Acos(Bx-C)+D \\ \\ \text{ And we can see that our function is of the form:} \\ \\ y=2cos(3x-0)-4=2cos(3x)-4 \\ \\ \text{ That is, we do not have a horizontal shift for this function. } \end{gathered}`

Therefore, we have to find the critical values from x = 0.

2. The period of the function is given by:

`\begin{gathered} \text{ Period =}(2\pi)/(B)=(2\pi)/(3) \\ \\ \text{ Period =}(2\pi)/(3) \end{gathered}`

Hence, the values we need to find go from 0 to (2/3)π.

3. Now, to find the critical values of the function, we have to find the derivative of the function, and then set the result to zero to find those values as follows:

`\begin{gathered} f(x)=2cos(3x)-4 \\ \\ f^(\prime)(x)=(2cos(3x)-4)^(\prime)=(2cos(3x))^(\prime)-(4)^(\prime)=2(cos(3x))^(\prime)(3x)^(\prime)-0 \\ \\ f^(\prime)(x)=2(-sin(3x))(3)=-2sin(3x)(3)=-6sin(3x) \end{gathered}`

Then we have:

`-6sin(3x)=0`

4. Now, we have to apply the inverse function of sine, the arcsine function, to solve the equation - but before, we can divide both sides by -6:

`\begin{gathered} (-6sin(3x))/(-6)=(0)/(-6)=0 \\ \\ sin(3x)=0 \\ \\ \sin^(-1)(sin(3x))=\sin^(-1)(0) \\ \\ 3x=0+2\pi n,3x=\pi+2\pi n \\ \\ \text{ Since sin\lparen x\rparen is zero in x = 0, and x = }\pi,\text{ before 2}\pi \end{gathered}`

5. Therefore, we have:

`\begin{gathered} 3x=0+2\pi n \\ \\ (1)/(3)(3x)=(1)/(3)(0+2\pi n)\Rightarrow x=(2\pi n)/(3)=(2)/(3)\pi n \\ \\ x=(2)/(3)\pi n\text{ \lparen General solution\rparen} \end{gathered}`

And also we have:

`\begin{gathered} \begin{equation*} 3x=\pi+2\pi n \end{equation*} \\ \\ (1)/(3)(3x)=(1)/(3)(\pi+2\pi n) \\ \\ x=(\pi)/(3)+(2)/(3)\pi n\text{ \lparen General solution\rparen} \end{gathered}`

6. Then, since the critical values are for one period starting at the horizontal shift, and the horizontal shift is 0, the values we need to find go from 0 to (2/3)π (one period of the function), then we have:

`\begin{gathered} x=(2)/(3)\pi n,x=(\pi)/(3)+(2)/(3)\pi n,\text{ and n is an integer \lparen in this case, n = 0, 1, 2, 3...\rparen} \\ \\ x=(2)/(3)\pi(0),(2)/(3)\pi(1),(2)/(3)\pi(2)\text{ \lparen this is not a value because it is greater than the period\rparen} \\ \\ x=0,(2)/(3)\pi\rightarrow\text{ \lparen These are two values\rparen} \end{gathered}`

And then we have the other possible values for the other general solution: