Answer:
f has no critical points.
Explanation:
We are given:
![f(x)=\cos(x)+2x](https://img.qammunity.org/2022/formulas/mathematics/high-school/4pgnts6sbnumtlzyy3pms9a7ho7gs42mpz.png)
A function has critical points whenever its derivative equals 0 or is undefined.
Differentiate the function:
![f'(x)=-\sin(x)+2](https://img.qammunity.org/2022/formulas/mathematics/high-school/w6u5bq9572kytsf5bvsz46ls04qnmvb9q3.png)
Since this will never be undefined, solve for its zeros:
![0=-\sin(x)+2](https://img.qammunity.org/2022/formulas/mathematics/high-school/aht9gsedwkpd6nnegqfy01dln9mw7pf4nv.png)
Hence:
![\displaystyle \sin(x)=2](https://img.qammunity.org/2022/formulas/mathematics/high-school/k0pkyc2x93o5io3fy7e534c34frfn1jpgk.png)
Recall that the value of sine is always between -1 and 1.
Thus, no real solutions exist.
Therefore, f has no critical points.