Question

How do you find out the intervals or inc/dec when the first derivative has no critical pts (no max/min)?

Answer 1

You can try taking the second derivative, and also plot points (find x- and y-intercepts, if possible.)

Answer 2

the function is xe^(-1/x^2)

Answer 3

i checked the graph online, and there is no concavity as well

Answer 4

\(\large xe^{-1/x^2} = \frac{x}{e^{x^{-2}}}\)

Answer 5

Did you try finding where the second derivative is zero?

Answer 6

umm, nop, let me try it now

Answer 7

the second dertivative seems like it's so hard to differentiate ..

Answer 8

It is a bit of a pain . .
Reason your way through it. You can't find any critical points - no relative maxima or minima, so there are no turning points; whatever it's doing, it's not changing direction, so if it is increasing anywhere, then it is increasing everywhere.

Plot some points and see if it is increasing or decreasing.

Answer 9

if the first derivative has no critical point, the curve is monotonically increasing or decreasing
it means the curve has no min/max point

Answer 10

oh ok, thanks.

to know if it is decreasing or increasing, so we just plug in points and try?

Answer 11

the best example is the line y=3x

Answer 12

the existence of a critical number in the first derivative does not guarantee an absolute max/min, just relative max/min

Answer 13

then how do i know if it is decreasing all the time or increasing all the time?
just plugging in pts?

Answer 14

Plugging in points seems a good way to go, try a couple negative values for x and a couple positive values for x and see what happens.

Answer 15

all right thx both!

Answer 16

i think there is a concavity..

Answer 17

equate each part of the fraction to zero

Answer 18

cuz my next part of the ques is asking about concavity

Answer 19

what about the concavity?

Answer 20

asking what is the concavity..that's means need to find out the 2nd derivative..which.. is ..so long and hard..
how can there be a concavity when there is no max and min..

Answer 21

if there is concavity, that means there should be a change of sign isnt it

Answer 22

There could be a stationary inflection point.

Answer 23

change of sign if the slope, not in the sign of the function

Answer 24

the second derivative is a measure of change in the slope, not the position of the point.

Answer 25

oh right, that's true , thx