# Pre-Calculus

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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)

• Pre-Calculus -

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

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