Question

Determine whether the function is concave up and where it is concave down. Also find all inflection points.

T(t) = 2t - t^3

Answer 1

can you find the first derivative (for the first step) ?

Answer 2

(just apply the power rule to each term, and you can use T'(t) as your notation for the left side)

Answer 3

T'(t) = 2 - 3t^2

Answer 4

then i have to set the function equals to zero?

Answer 5

no, you have to find the second derivative, and only then set it to equal zero.

Answer 6

T'(t) = -6t

Answer 7

yes, but you meant,
T''(t)=-6t

Answer 8

-6t = 0
t = 0?

Answer 9

since 0/-6 = 0

Answer 10

\(\normalsize\color{blue}{ T\prime \prime(t)=-6t }\)

Answer 11

yes, t=0 is (the only) inflection point.

Answer 12

when the \(\normalsize\color{blue}{ T\prime \prime(t) }\) is greater than zero, for what values of \(\normalsize\color{blue}{ t }\)?

Answer 13

idk...

Answer 14

you have \(\normalsize\color{blue}{ T\prime \prime(t)=-6t }\)
what happens when t is negative, what happens when t is positive?

Answer 15

if \(\normalsize\color{blue}{ t }\) is a positive number, (starting from any value that is greater than zero),
then the \(\normalsize\color{blue}{ T\prime \prime(t) }\) is less than zero. Right?

Answer 16

ye

Answer 17

What can you say about \(\normalsize\color{blue}{ T\prime \prime(t) }\)
when \(\normalsize\color{blue}{ t }\) is a negative number?

Answer 18

T''(t) is greater than zero

Answer 19

Yes.

Answer 20

\(\normalsize\color{blue}{ T\prime \prime(t)=-6t }\)

So:
If, \(\normalsize\color{blue}{ t>0 }\) then, \(\normalsize\color{blue}{ T\prime \prime(t)<0 }\)
AND
If, \(\normalsize\color{blue}{ t<0 }\) then, \(\normalsize\color{blue}{ T\prime \prime(t)>0 }\)

Answer 21

can you give me the concavity in interval notation?

Answer 22

concave down and concave up?

Answer 23

yes, one interval for concave up, and the other is concave down. (I have even asked the question in the needed for you order :) )

Answer 24

I'll give you two intervals, and you will tell me which is concave up, and which is concave down, okay?

Answer 25

ok

Answer 26

1) \(\normalsize\color{blue}{ (-\infty,0) }\)
2) \(\normalsize\color{blue}{ (0,+\infty) }\)

Answer 27

1. concave up
2. concave up?

Answer 28

can't be that both are concave up.

Answer 29

T''(t) > 0 = concave up
T''(t) < 0 = concave down

Answer 30

yes, but on what interval is T''(t)>0 and on what interval is T''(t)<0 ?

(reminding that T''(t)=-6t in our case)

Answer 31

\[(-\infty, 0) concave down ||(0,+\infty) concave up\]

Answer 32

it is the other way, and also you can use ~ for a space.

Answer 33

how would i sketch the graph for this?

Answer 34

I wouldn't just sketch it based on the concavity. although you can make a sketch something like