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Determine the open intervals on which the graph of f(x) = x - 4cos x is concave upward or concave downward

Respuesta :

frika

The graph of the function [tex]f(x)[/tex] is

  • concave upward, when [tex]f''(x)>0;[/tex]
  • concave downward, when [tex]f''(x)<0.[/tex]

Find [tex]f''(x):[/tex]

1.

[tex]f'(x)=(x-4\cos x)'=1+4\sin x;[/tex]

2.

[tex]f''(x)=4\cos x.[/tex]

Now:

1. when [tex]4\cos x>0,[/tex] the graph of the function is concave upward and this is for

[tex]x\in \left(-\dfrac{\pi}{2}+2\pi k,\dfrac{\pi}{2}+2\pi k\right), \text{ where } k\in Z.[/tex]

2.  when [tex]4\cos x<0,[/tex] the graph of the function is concave downward and this is for

[tex]x\in \left(\dfrac{\pi}{2}+2\pi k,\dfrac{3\pi}{2}+2\pi k\right), \text{ where } k\in Z.[/tex]