# Explain why the graph of f(x)=(3x−π)cos(1/2x) lies below the x-axis for −π<x<1/3π , above the x-axis for 1/3π <x<π

*print*Print*list*Cite

You need to solve the equation f(x)=0 such that:

`(3x−pi)cos(x/2) = 0`

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

`cos (x/2) = 0 =gt x/2 = pi/2 =gt x = pi`

You need to select a value for x, larger than `pi/3` and smaller than `pi ` such that:

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

You need to substitute `2pi/3` for x in equation `(3x−pi)cos(x/2) ` such that:

`(2pi−pi)cos(pi/3) = pi*(1/2) = pi/2 gt 0`

You need to select a value for x, larger than `pi` such that:

`x = pi+pi/3 = 4pi/3`

You need to substitute `4pi/3` for x in equation `(3x−pi)cos(x/2)` such that:

`(4pi−pi)cos(2pi/3) = -pi*(1/2) = -pi/2lt 0`

Hence, if `x in (pi/3,pi), ` then the values of function are positive, thus the graph of function is above x axis and if `x in (-pi,pi/3), ` the values of function are negative, thus the graph of function is below x axis.

Substituting -3.14 for `pi` , 1.04 for `pi/3` and 3.14 for `pi` , you may notice that if `x in (-3.14 , 1.04)` , the red curve goes below x axis and if `x in (1.04 , 3.14)` , the red curve goes above x axis.