Question

find the min of...
g(t)=t^3-9t+5 on interval [-1.5, 3]

Answer 1

why isnt it 3?

Answer 2

don't i just take the derivative of whats given, set it to zero, and solve?

Answer 3

g(t)=t^3-9t+5 on interval [-1.5, 3]

g'(t)= 3t² - 9 => 3(3t - 3) => t = 1

Answer 4

That's the first step, isn't it?

Answer 5

oops i made a mistake

Answer 6

3(t² - 3) = 0 => t = + sqrt3 and t = -sqrt 3

Answer 7

did you do that?

Answer 8

yo!

Answer 9

is that right, saifoo?

Answer 10

i dont think it is

Answer 11

\[\checkmark\]

Answer 12

Melind is correct

Answer 13

i must not be putting it in my hw site correctly then...

Answer 14

Now i'd probably do like this. Considering sqrt3 is 1,7

t -1,7 ... 1,5 ... 3
g'(t)
g(t)

and find the values and signs (positive and negative)

Answer 15

im confused

Answer 16

me too, kinda lol wait a sec pls

Answer 17

ok :)

Answer 18

i guess we should draw a graph to see it clearer

Answer 19

@KingGeorge hey you there?

Answer 20

So you found the critical points at \(t= \pm \sqrt3\). Next step, take the second derivative. Then you get\[f''(t)=6t\]Since this is negative when \(t=-\sqrt3\), that must be where the minimum is.

Last step, solve for \(f(-\sqrt3)\).

Answer 21

That was a mistake. We're looking for where the second derivative is positive, not negative, so the minimum would be at \(t=+\sqrt3\). So solve for that instead of \(-\sqrt3\).

Answer 22

You should get \[f(\sqrt3)=3\sqrt3 -9\sqrt3+5=-6\sqrt3+5\]

Answer 23

you forgot the squared

Answer 24

Now we need to check the boundaries. If we try \(t=-1.5\), we get a value greater than \(f(\sqrt3)\), and if we try \(t=3\), we also get a value greater than \(f(\sqrt3)\). So \(-6\sqrt3+5\) is the minimum value.

Answer 25

The function is \[t^3-9t+5\] correct? In my calculation above I've already cubed \(t\)

Answer 26

nvm then

Answer 27

Thanks :D
I posted another Max/Min prob if you wanna help with that too.

Answer 28

You're welcome.

Answer 29

King is awesome, thanks.

Answer 30

Why thank you. :)