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Let f(x)=4x3+1. Fin… - QuestionCove
OpenStudy (anonymous):

Let f(x)=4x3+1. Find the open intervals on which f is increasing (decreasing). Then determine the x coordinates of all relative maxima (minima). 1. f is increasing on the intervals 2. f is decreasing on the intervals 3. The relative maxima of f occur at x = 4. The relative minima of f occur at x = Notes: In the first two, your answer should either be a single interval, such as (0,1), a comma separated list of intervals, such as (-inf, 2), (3,4), or the word "none".

3 years ago
OpenStudy (anonymous):

it is \[4x ^{3}+1\]

3 years ago
OpenStudy (anonymous):

@aum

3 years ago
OpenStudy (anonymous):

@sourwing should it start with factorization?

3 years ago
OpenStudy (aum):

f(x) = 4x^3 + 1 f'(x) = 12x^2 12x^2 is always positive (except at x = 0). Therefore, f(x) is an increasing function at all x except x = 0.

3 years ago
OpenStudy (aum):

2. There is no interval in which f(x) decreases. None.

3 years ago
OpenStudy (anonymous):

so the interval would be (-inf,0)U(0,inf)?

3 years ago
OpenStudy (aum):

Strictly speaking, yes. But the instruction says: comma separated intervals.

3 years ago
OpenStudy (aum):

1. (-inf, 0), (0, inf) 2. none

3 years ago
OpenStudy (anonymous):

ok so what about 3 and 4?

3 years ago
OpenStudy (aum):

If the function is always increasing, with its range being (-inf, inf) there is no relative maxima or minima.

3 years ago
OpenStudy (anonymous):

m getting ! wrong for some reason

3 years ago
OpenStudy (anonymous):

sorry I meant 1

3 years ago
OpenStudy (aum):

Then try (-inf, inf). Strictly speaking it is not increasing at x = 0, but it is not decreasing either. It is called a stationary point. But some times they group it under increasing part.

3 years ago
OpenStudy (anonymous):

it went through this time

3 years ago
OpenStudy (anonymous):

thats weird

3 years ago
OpenStudy (aum):

Some books define increasing as parts where the function is non-decreasing. If that is the definition, then at x = 0 the function is not decreasing and therefore can be grouped with the increasing part.

3 years ago
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