Question

I need help with the average rate of change f(t)=9/t ,t=a, t=a+h

Answer 1

\[f(t) = \frac{ 9 }{ t}\]

\[(t = a) | t = (a+h)\]

Answer 2

what approach comes to mind?

Answer 3

f(b)-f(a)/b-a

Answer 4

do you know how to take the power rule of differentiation? this is average rate so i'm thinking slope here

Answer 5

no im not sure

Answer 6

another way is this, I guess we could use the definition of a limit here.

\[\lim_{h \rightarrow 0} (\frac{ f(a)-f(a+h) }{ a-(a+h) })\]

Answer 7

yeah that is the way im talking about

Answer 8

im not sure how to substitute this ?

Answer 9

\[\frac{ \frac{ 9 }{ a } - \frac{ 9 }{ a+h }}{ h } \]

Answer 10

is it like \[f(9/a+h)-f(9/a)/(a+h)-9/a\]

Answer 11

i thought b went before a ?

Answer 12

you are correct i'm wrong

Answer 13

it would be in the form f(a+h)-f(a)

Answer 14

\[\frac{ f(\frac{ 9 }{ a+h })-f(\frac{ 9 }{ a }) }{ h }\]

Answer 15

is this difference quotient?

Answer 16

@Nnesha clarification?

Answer 17

would the x values at the bottom just be (a+h)-(a) @Nnesha

Answer 18

what happens to 9/t?

Answer 19

my assumption is t is substituted where a is t = a

Answer 20

yeah

Answer 21

what x value ??
btw riss is it for calculus class ?

Answer 22

this is for precalc

Answer 23

question 8

Answer 24

I need help with this. I know that D and E are correct because there is a infinite discontinuity at 1 and also a vertical asymptote. I'm thinking that C is also not the right answer. Can someone help me? Please and thank you!

Answer 25

ahh i see what you were talking about
and in this case x is t so a+h
but.. the denominator would not be a+h - a
we have to find the common denominator

Answer 26

rissa i can't open that file
have you started the limit types q in class yet or not ?

Answer 27

ooop i put my question in your thing. soooo sorry

Answer 28

ive been taking a summer course but i caught bronchitis and then i couldnt be in the class. so they moved my final however the final is suppose to be for fall and spring students which im not. so there are alot of things i dont know on this exam.

Answer 29

Determine the average rate of change of the function f(t) = 9
t
between the values t = a and t = a + h.
Simplify your answer.

Answer 30

9/t

Answer 31

alright so i don't think we need to take the limit looks like simple algebra question \[\large\rm \frac{ y_2 -y_1}{ x_2 -x_1 }=average ~rate ~of~change\]
in this question x=t
so substitute t values into the function to find Y_1 and y_2

Answer 32

as mentioned above n Photon336 comment replace `t` with a and `a+h`
\[ \frac{ f(\frac{9}{a+h}) -f(\frac{9}{a} )}{ t_2 - t_1 }\]
t_2 = a+h
and t=a
\[ \frac{ f(\frac{9}{a+h}) -f(\frac{9}{a} )}{ \color{Red}{a+h} - \color{blue}{a} }\]

Answer 33

actually i just figured out that is not the correct way to right this
let me correct that it should be \[\frac{ f(\color{Red}{a+h})-f(\color{blue}{a}) }{ \color{Red}{a+h}-\color{blue}{a} }\]
at the denominator f(a+h) means t=a+h if a+h is input what would be the output ?

Answer 34

\[f(t)=\frac{9}{t}\] replace that `t` with a what would you get ??
and then replace that `t` with `a+h` what would you get ?

Answer 35

9/a

Answer 36

\[(9/a+h)-(9/a) /a+h-9/a\]

Answer 37

input would be at the denominator
so t values would be at the denominator a+h-a like this

Answer 38

\[f(a+h)\] means if input is a+h what would be the output simply
substitute t with `a+h` into this function \[\rm f(t)=\frac{9}{\color{Red}{t}}\]
\[\rm f(\color{ReD}{a+h})=\frac{9}{\color{Red}{a+h}}\]

Answer 39

now replace f(a+h) with 9/a+h

Answer 40

im trying to send you a photo

Answer 41

of where im at

Answer 42

f(a) when input is a the output would be what?
to find out replace t with a into the function 9/t \[\rm f(t) =\frac{9}{t} \]
\[\rm f(\color{Red}{a}) =\frac{9}{\color{Red}{a}} \]
plugin
\[\frac{ y_2 -y_1 }{ x_2 -x_1 } = \frac{ \frac{9}{a+h}-\frac{9}{a} }{ a+h-a }\]

Answer 43

how did you get 9/a at the denominator??

Answer 44

i thought the formula was f(b)-f(a)/b-a

Answer 45

i got 9/a because i thought you replace t with a

Answer 46

yeah replace t with a into the original function
f(t)=9/t

Answer 47

f(b)-f(a) over b-a is same as y_2 - y_1 over x_2 -x_a
b and a are the input value

Answer 48

im so confused :(

Answer 49

is there anyway that you can send me a photo of the set up and how you answered it?

Answer 50

i already posted the half of the solution

Answer 51

im just confused about the bottom

Answer 52

\[\rm \frac{ \frac{9}{a+h}-\frac{9}{a} }{ a+h-a }\]


if t=a then f(a) = 9/a
if t=a+h then f(a+h)=9/a+h
\[\rm \frac{ \color{Red}{\frac{9}{a+h}-\frac{9}{a} }}{ a+h-a }\]
simplify the numerator first
find the common denominator

Answer 53

omg I GOT it!! im so sorry it took me so long to figure this out.