Calc 1 Critical Numbers?

Question

I thought there would be more crit numbers for this. Can
anyone help me get the right answer . Due tommorow


Find the Critical Numbers and find the max min.
Then find the second derivative

y = x+cosx



i got the first deriv as 1 - sinx or something

Answer 1

y = x + cosx
y' = 1 - sinx
y" = - cosx

1 - sinx = 0
sinx = 1
x = π/2 ± 2nπ

Answer 2

use the first derivative to get your critical #s. i.e. 1 - sin x = 0

find the critical values at the different points. i.e. plug your critical #s back into the equation.

finally, the 2nd derivative is y"= -cos x
because d(1)=0 and the d(sinx)=cos x

Answer 3

first derivative is 1-sinx
the critical points when 1-sinx=0
hence the critical points when x= (4n+1)pi/2
where n = 0,1,2....

there is no max and min as it tends to +/- infinity

the second derivative is -cosx