Answer:
f has no critical points.
Step-by-step explanation:
We are given:
[tex]f(x)=\cos(x)+2x[/tex]
A function has critical points whenever its derivative equals 0 or is undefined.
Differentiate the function:
[tex]f'(x)=-\sin(x)+2[/tex]
Since this will never be undefined, solve for its zeros:
[tex]0=-\sin(x)+2[/tex]
Hence:
[tex]\displaystyle \sin(x)=2[/tex]
Recall that the value of sine is always between -1 and 1.
Thus, no real solutions exist.
Therefore, f has no critical points.
