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Sagot :
x-coordinates for the maximum points in any function f(x) by f'(x) =0 would be x = π/2 and x= 3π/2.
How to obtain the maximum value of a function?
To find the maximum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.
we want to find x-coordinates for the maximum points in any function f(x) by f'(x) =0
Given f(x)= 4cos(2x -π)
[tex]f'(x) = 0\\- 4sin(2x -\pi ) =0\\\\sin (2x -\pi ) =0 \\2x -\pi = k\pi ... k in Z[/tex]
In general [tex]x=(k+1)\pi /2[/tex]
from x = 0 to x = 2π :
when k =0 then x = π/2
when k =1 then x= π
when k =2 then x= 3π/2
when k =3 then x=2π
Thus, X-coordinates of maximum points are x = π/2 and x= 3π/2
Learn more about maximum of a function here:
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