f'=-sinx-.26=0
solve for x
f"=-cosx
Find all relative extrema. Use the Second Derivative Test where applicable.
f(x)= cosx - x (0,2ð)
2 answers
f' = -sinx-1
f'=0 at sinx = -1, so x = 3π/2
f" = -cosx, so f"(3π/2) = 0
Looks like an inflection point, not max or min.
So, no extrema anywhere.
f'=0 at sinx = -1, so x = 3π/2
f" = -cosx, so f"(3π/2) = 0
Looks like an inflection point, not max or min.
So, no extrema anywhere.