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...
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` )