Question

locate all the critical points of s(t)=(t-1)^2(t+6)^3

Answer 1

I assume you are suppose to use calculus?

Answer 2

yes

Answer 3

To acquire the critical points take the derivative

set y=0,
solve for x,

If any of the values of x are not in the domain of the original function ignore them

Then,
check the domain of the derivative if you find numbers that aren't in the domain of the derivative but are in the domain of the original function they are critical numbers.

If any of the values of x are not in the domain of the original function ignore them

Answer 4

Do that and I will show you the next step

Answer 5

well there isn't really a next step but I will show you how to draw the function from the critical numbers

Answer 6

wait im confused can you show me all the steps? i dont understand

Answer 7

well take the derivative first

Answer 8

im really bad... 2(t-1)^1*2(t+6)^3+6(t+6)*(t-1)? is thar the start of it?

Answer 9

do you want me to show you the easy way to do derivatives?

Answer 10

I will just show you my method it is fail proof

Answer 11

sure

Answer 12

You better be grateful haha, this will take a little bit of time

Answer 13

:) thanks

Answer 14

s(t)=(t-1)^2(t+6)^3

so first split the function into smaller functions and take the derivative
so we are left with,

g(t) = (t-1)^2
and
q(t) = (t+6)^3

FIRST PART

g(t) = (t-1)^2
We need to use chain rule to figure out what g'(t) is

so we simply split this into smaller functions

m(t) = t^(2)
m'(t) = 2t

l(t) = t-1
l'(t) = 1

Now we apply chain rule which is simply (MEMORIZE THIS RULE)
g'(t) = m't( l(t) )*l't

so after subbing in according to this rule we are left with

g(t) = (t-1)^2
g'(t) = 2(t-1)*1

Now we are half way there

PART B

q(t) = (t+6)^3

same deal, split them

u(t) = t^3
u'(t) = 3t^2

o(t) = t + 6
o'(t) = 1

Now chain rule again,

q'(t) = u'(o(t))*o'(t)
q'(t) = 3(t+6)^2 * 1

so we have

q(t) = (t+6)^3
q'(t) = 3(t+6)^2 * 1

Final Part, putting it together

g(t) = (t-1)^2
g'(t) = 2(t-1)

q(t) = (t+6)^3
q'(t) = 3(t+6)^2

we clearly need to use product rule here so simply just use the rule

s'(t) = g'(t)q(t) + q'(t)g(t)
s'(t) = 2(t-1)(t+6)^3 + 3(t+6)^2(t-1)^2


so we are left with

s(t)=(t-1)^2(t+6)^3
s'(t) = 2(t-1)(t+6)^3 + 3(t+6)^2(t-1)^2

Answer 15

It looks long and time consuming but this is the best way to do derivatives, just keep doing them like this it will get very systematic and eventually you will be able to do them in your head really fast

Answer 16

Using this method you have no excuses for making a mistake when taking derivatives ha

Answer 17

wow thats long lol

Answer 18

I made a mistake
g'(t) = m't( l(t) )*l't

should be,

g'(t) = m'( l(t) )*l'(t)

Answer 19

Yes it is long but you need to see all the steps so you understand the mental process

Answer 20

yea im saving it lol

Answer 21

how do we find the critical points now?

Answer 22

Finding critical points

Step 1

Check domain of s'(t) and s(t) (re-read what I wrote about this above)
Both are R so don't worry about it

Step 2
(Good to write this on your test or assignment!!!)
set s'(t) = 0

so,

0 = 2(t-1)(t+6)^3 + 3(t+6)^2(t-1)^2

solve for t, what ever you get as t are your critical points

Answer 23

But yeah, I cant stress this enough do derivatives the long way until you are super comfortable with them, even then it doesn't hurt to do them the long way. Once you do it this way you will never forget how to do them, as there are only like 3 formulas you need to remember.

Answer 24

can you help me solve it, im getting mixed up with the numbers and exponents

Answer 25

Yeah this looks painful to expand

Answer 26

Are you in intro to calculus?

Answer 27

or is your prof just evil

Answer 28

i just have an evil professor lol that doesnt explain anything