Question

For position function s(t):=20-5t^2, if h is a small number, find the average rate of change of s(t) on the interval [2,2+h]. (Simplify fully)

Answer 1

ug ..... i spose

Answer 2

?

Answer 3

s(2+h) -s(2)
-----------
h

Answer 4

its alot of work just to prove that derivatives work ....

Answer 5

lol true.

Answer 6

20-5(2+h)^2 -(20-5(2)^2)
------------------------
h


20-5(4 +4h +h^2) -20 +20
------------------------
h


20-20-20h-5h^2
----------------
h


-20h-5h^2
---------- = -20 -5h
h

Answer 7

I think I have to use the formula (f(a)-f(x))/(a-x)

Answer 8

Can you explain to me how to use that formula?

Answer 9

now for the derivative we get:

-10t ; at t=2 we get -20

when h=0, we get -20

Answer 10

its the same formula

Answer 11

oh no derivatives yet

Answer 12

can you explain to me how to use it? like what's a and what's x?

Answer 13

on the interval

Answer 14

yes, you are doing the proof for derivatives so that in the next few weeks your teacher will say, and this is how we do it the easy way...

Answer 15

haha ok.

Answer 16

f(a) and f(x) simply mean to use the formula with an a and an x instead of a t like this:


f(a) - f(x)
(20-5a^2) - (20-5x^2)
-------------------
a-x

Answer 17

20-5a^2 -20+5x^2
-----------------
a - x

-5a^2+5x^2
-------------
a - x


-5(a^2-x^2)
-------------
a - x


-5(a+x)(a-x)
-------------
a - x

-5(a+x)

Answer 18

ok, what do you plug into f(a)?

Answer 19

in [2,2+h]

Answer 20

.............. a = 2 and x = h

Answer 21

?

Answer 22

is a=2 from the 2+h part or from just the 2 part.

Answer 23

it doesnt matter..... a =2 OR a=h, OR x=2 OR x=h....

-5(a+x)

-5(2+h) is the same things as -5(h+2)

Answer 24

I'm confused.

Answer 25

Like, I know the forumla, I just don't understand how to use it.

Answer 26

then dont use it lol

Answer 27

does it matter if f(a) = 2 and f(x) = (2+h)? no

Answer 28

becasuse when h=0 it just disappears to -20

Answer 29

\[\frac{y-y_0}{x-x_0}\] do you understand this formula?

Answer 30

yes.

Answer 31

that is all it is; but they rename that parts ..... but it doesnt matter what they name them, they still do the same things

Answer 32

so the bottom part is a-x... don't you put 2-h?

Answer 33

why did you only put h?

Answer 34

\[\frac{f(a)-f(x)}{a-x} \iff \frac{y-y_0}{x-x_0} \iff \frac{f(x+h)-f(x)}{h}\]

Answer 35

oh ok.

Answer 36

because \((x+h)-(x)=h\)

Answer 37

why?

Answer 38

why what? why does x-x+h = 0+h = h?

Answer 39

oh ok. just kidding.

Answer 40

\[(x+h)-(x)\]
\[x+h-x\]
\[x-x+h\]
\[(x-x)+h\]
\[0+h\]
\[h\]

Answer 41

so the formula would be \[(f(2+h)-f(2))\div(h)\]

Answer 42

yes

Answer 43

ok.

Answer 44

so its ((20-5(2+h)^2)-(20-5(2)^2))/h

Answer 45

yes

Answer 46

so (20-5(2+h)^2)/h

Answer 47

and finally -20-h?

Answer 48

yes

Answer 49

:) Thank you