Question

Let f(x) = 6x^{3}+7. 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 =

Answer 1

so this is a cal question?

Answer 2

yes it is. calc 1

Answer 3

ok first step to get any of this is to differentiate f

Answer 4

\[f(x)=6x^{3}+7\]
so
\[f'(x)\]

Answer 5

there is suppose to be =? after the f'(x) there

Answer 6

f'(x) = 18x^2

Answer 7

now
if f'>0 then f is increasing
if f'<0 then f is decreasing

Answer 8

when f'=0 we have f is doing neither
and f'=0 gives us our critical numbers

Answer 9

now what do you know about the factor x^2?
Is x^2 ever negative?

Answer 10

it is never negative

Answer 11

\[\text{ since } 18x^2 \ge 0 \text{ for all x} \\ \text{ then this tells us } \text{ f is never descreasing }\]

Answer 12

since f' is never negative

Answer 13

anyways 18x^2=0 when x=?

Answer 14

x = 0

Answer 15

so that means everywhere but at x=0 f is increasing
since f'=18x^2>=0 for all x

Answer 16

also a function that is never decreasing doesn't have any relative min or max

just like a function that is never increasing doesn't have any relative min or max

for there to be a min or max at a critical number the function must switch from decreasing to increasing there or increasing to decreasing

Answer 17

that is for a relative min/ max anyways

Answer 18

okay so there is no relative min or max, nor is the function have any decreasing intervals. So it only has an increasing interval at (-inf,inf). Is this correct?

Answer 19

well i would say isn't increasing at x=0

Answer 20

because f'=0 at x=0

Answer 21

but it is increasing everywhere else

Answer 22

(-inf,0) U (0,inf)

Answer 23

Okay, thanks for the help.

Answer 24

np