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Given the function f'(x) = x³ - 3x,
determine all intervals on which f is increasing.

User Michael Macha
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1 Answer

18 votes
18 votes

Explanation:

f(x) increases if and only if f'(x) is positive.

so, we need to find all intervals, where f'(x) is positive.

for this we need to find the zeroes of f'(x).

one is clear : x = 0.

and we can do a factorization for f'(x) based on this :

f'(x) = x(x² - 3)

the other 2 zeroes are given by

x² - 3 = 0

x² = 3

x = ±sqrt(3)

so, we have the intervals

(-infinity to -sqrt(3)]

(-sqrt(3) to 0)

[0 to +sqrt(3)]

(+sqrt(3) to +infinity)

for x in (-infinity, -sqrt(3)] f'(x) is not positive (max. 0), so f(x) is not increasing.

for x in (-sqrt(3), 0) f'(x) is positive, so f(x) is increasing.

for x in [0, +sqrt(3)] f'(x) is again not positive (max. 0), so f(x) is not increasing.

for x in (+sqrt(3), +infinity) f'(x) is positive, so f(x) is increasing.

please consider the brackets.

"(" and ")" mean the corresponding limit of the interval is not included.

"[" and "]" mean the corresponding limit of the interval is included.

as f'(x) = 0 means the function is neither increasing nor decreasing at this point. there is a (local) max or min at that point. and the tangent at that point is horizontal (slope 0).

so, for the increasing parts of f(x) we cannot include these points.

User Laurent Brieu
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2.7k points