Question

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".

Answer 1

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

Answer 2

@aum

Answer 3

@sourwing should it start with factorization?

Answer 4

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.

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

ok so what about 3 and 4?

Answer 10

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

Answer 11

m getting ! wrong for some reason

Answer 12

sorry I meant 1

Answer 13

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.

Answer 14

it went through this time

Answer 15

thats weird

Answer 16

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.