Explain why the graph of f(x)=(3x−π)cos(1/2x) lies below the x-axis for −πI get that for x the interval [pi/3,pi], 3x-pi lies in the interval [0,2pi] so 3x-pi is positive. Also for x in the...

Explain why the graph of f(x)=(3x−π)cos(1/2x) lies below the x-axis for −πI get that for x the interval [pi/3,pi], 3x-pi lies in the interval [0,2pi] so 3x-pi is positive. Also for x in the interval [pi/3,pi] 1/2(x) lies in the interval [pi/6,pi/2] so cos(1/2(x)) is positive. Hence the graph of y= (3x-pi)cos(1/2(x)) is above the x axis on the interval [pi/3,pi]. For x in the interval [-pi,pi/3] 3x-pi lies in the interval [-4pi,0] so 3x-pi is negative. Also for x in the interval [-pi,pi/3] 1/2(x) lies in the interval [-pi/2,pi/6]. . . . Pi/6 is positive not negative! help!

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to solve the equation `(3x−pi)cos(x/2) = 0`  such that:

`3x - pi = 0 =gt 3x = pi =gt x = pi/3`

`cos (x/2) = 0 `

You need to remember that the values of cosine function rest positive in 1 and 4 quadrants, such that:

`x/2 = arccos 0 =gt x/2 = pi/2 =gt x = pi`

`x/2 = 3pi/2 =gt x = 3pi`

You need to substitute x by a value in interval `[pi;3pi]`  moving in positive direction over trigonometric quadrants.

You need to notice that the graph of function intercepts axis at `x=pi/3 in [pi,3pi]`

You need to consider a value `x = pi - pi/3 = 2pi/3`  such that:

`3x - pi = 2pi - pi = pi`

`cos 2pi/6 = cos pi/3 = 1/2`

You need to consider a value `x = pi+ pi/3 = 4pi/3`  such that:

`4pi-pi = 3pi`

`cos 4pi/6 = cos 2pi/3 = -cos pi/3 = -1/2`

Notice that if `x in [pi,3pi/2]` , the values of function are negative (for `x=4pi/3 =gt f(x) = -3pi/2` ), hence the graph of function goes below x axis (consider `pi~~3.14` )

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