if f' > 0, f is increasing. So,
f(x)=3
f' = 0
f is constant everywhere
f(x) = x^(2/3)
f' = 2/3 x^(-1/3)
f' > 0, so f is increasing everywhere
f(x) = -x^(3/4)
f' = -3/4 x^(-1/4)
f' < 0 so, f is decreasing everywhere
rather poor examples, since there are no finite open intervals involved
Determine the open intervals on which the function is increasing, decreasing, or constant.
f(x) = 3
f(x) = x^(2/3)
f(x) = -x^(3/4)
1 answer
1 answer