Question

screenshots attached!

Answer 1

Average rate of change on the interval [a,b] for f(x)

\[\Large m = \frac{f(b)-f(a)}{b-a}\]

m is the average rate of change

Answer 2

the part where it says solve on this interval confuses me, i don't know how to get that

Answer 3

thank your for taking the time to help me, greatly appreciate it

Answer 4

we have

f(t) = 4t^2 - 1

Answer 5

what is f(5) ?

Answer 6

the domain?

Answer 7

no what happens when you plug t = 5 into f(t) = 4t^2 - 1

Answer 8

f(t) = 4t^2 - 1

f(5) = 4(5)^2 - 1 ... replace every 't' with 5

f(5) = ???

Answer 9

oh 99!!!!

Answer 10

what is f(5.1) equal to

Answer 11

5.1 or 5? dont i plug in 5

Answer 12

when we say f(5.1) and when f(t) = 4t^2 - 1, we are replacing every t with 5.1

Answer 13

f(3) means replace every t with 3

etc etc

Answer 14

4(5.1)^2-1=103.04

Answer 15

so we know this

f(5) = 99
f(5.1) = 103.04

Answer 16

we now plug that into the formula

\[\Large m = \frac{f(b)-f(a)}{b-a}\]

in this case

a = 5
b = 5.1

Answer 17

so...

\[\Large m = \frac{f(b)-f(a)}{b-a}\]

\[\Large m = \frac{f(5)-f(5.1)}{5-5.1}\]

\[\Large m = \frac{99-103.04}{5-5.1}\]

\[\Large m = \frac{-4.04}{-0.1}\]

\[\Large m = 40.4\]

Answer 18

perfect!!! thank you!!!!!

Answer 19

oh sry I have a = 5.1 and b = 5 but it really doesn't matter (since you can swap a and b easily)

Answer 20

anyways, the average rate of change is 40.4

this represents the slope of the line through (5, 99) and (5.1, 103.04)

Answer 21

so my final 3 answers? i don't get how the web assign system wants it

Answer 22

so that's why it looks very similar to the slope formula (because in a way, it is the slope formula)

Answer 23

do you know how to find the derivative?

Answer 24

oh genius!

Answer 25

no! i was having a lot of trouble yesterday with that

Answer 26

have you learned about calculus topics yet? like limits and derivatives

Answer 27

i am taking this class online and I've tried, but its harder online

Answer 28

technically according to my homework i should know them already...

Answer 29

so the lesson plan isn't that nailed down

Answer 30

not at all!!

Answer 31

ok well when you get free time, I recommend you read over these notes

http://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx

go through it and see what makes sense and what doesn't. Then you can come back here and ask questions on stuff you don't get

Answer 32

anyways, you'll use derivatives to derive f(t) to get f ' (t)

this will give you a function that allows you to find the instantaneous rate of change at any point

Answer 33

thank you, the link has been saved, and will look at it after i finish these problems!

Answer 34

okay

Answer 35

to derive, we will use the power rule

http://tutorial.math.lamar.edu/Classes/CalcI/DiffFormulas.aspx

http://www.mathsisfun.com/calculus/derivatives-rules.html

which is

if f(x) = x^n then f ' (x) = n*x^(n-1)

Answer 36

and we'll also use the constant coefficient rule

Answer 37

so

f(t) = 4t^2 - 1

d/dt [f(t]) = d/dt[4t^2 - 1]

f ' (t) = 2*4t^(2-1)

f ' (t) = 8t

Answer 38

im familiar with the power rule, i just always apply it at worng times!

Answer 39

okay.

Answer 40

what is f ' (5) equal to?

Answer 41

thank you for all the links!

Answer 42

99?

Answer 43

f ' (t) = 8t

f ' (5) = 8*5

f ' (5) = 40

Answer 44

so the instantaneous rate of change at t = 5 is 40

Answer 45

notice how it's pretty close to that average rate of change we got before, which was 40.4

Answer 46

yes!!!!! i see it clearly, but without help i won't figure it out

Answer 47

you will once you practice more and read over notes like what I posted

Answer 48

thank you! i will!

Answer 49

you're welcome

Answer 50

so then my answers are 40 at t5 and 40.1 at t5.1

Answer 51

not 40.1

Answer 52

40.1 goes on the top

Answer 53

f ' (t) = 8t

f ' (5.1) = 8*5.1

f ' (5.1) = 40.8

Answer 54

the instantaneous rate of change at t = 5.1 is 40.8

Answer 55

and the top answer? the first empty box?

Answer 56

that was the avg rate of change, aka the slope of the line through (5, 99) and (5.1, 103.04)

Answer 57

this is what I get

Answer 58

i thought that one is 40.4?

Answer 59

the last one, you'll do

V(s) = s^3

V ' (s) = 3s^2

V ' (19) = 3*19^2

V ' (19) = 1083

Answer 60

this is what i have

Answer 61

correct so far

Answer 62

of course because you gave them to me! lol

Answer 63

thank you for the second question:)

Answer 64

lol well hopefully it's clicking that you could find similar problems in the future

Answer 65

np

Answer 66

I'm going to practice different versions after i get this one correctly

Answer 67

good idea

Answer 68

i think I'm done?

Answer 69

no its .8?

Answer 70

the only one left is we just subtract 40.8-40

Answer 71

see this

Answer 72

\[\Large m = \frac{f(b)-f(a)}{b-a}\]

\[\Large m = \frac{f(5)-f(5.1)}{5-5.1}\]

\[\Large m = \frac{99-103.04}{5-5.1}\]

\[\Large m = \frac{-4.04}{-0.1}\]

\[\Large m = 40.4\]

Answer 73

oh gosh i was so wrong!!!

Answer 74

thank you it was all right!

Answer 75

do u want to help me with my last one? I'll really try to do it?

Answer 76

ok go ahead and post it

Answer 77

:) Thank you

Answer 78

we'll use the idea that