Question

Using your function f(x) from question 2. Demonstrate and explain how to find the average rate of change between year 3 and 5, and between year 5 and 7. Explain what the average rate of change represents to the frog population.

Answer 1

\[f(x) = A * (B)^x\]

Answer 2

@ranga

Answer 3

@BigTeach2345 Think you can help?

Answer 4

@AllTehMaffs @ganeshie8 @shamil98 @hartnn

Answer 5

assign some values to A and B

Answer 6

initial frog population = A = 10 ?
rate of change of population = B = 2 times each year ?

Answer 7

A=10
B=2

Answer 8

f(x)=10*2^x

Answer 9

looks good !
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Demonstrate and explain how to find the average rate of change between year 3 and 5,
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Answer 10

wat do u knw about average rate of change ?

Answer 11

the slope stays the same right?

Answer 12

yes

average rate of change of \(f(x)\) between \(x = a\) and \(x = b\) is :-

\(\huge \frac{f(b) -f(a)}{b-a}\)

Answer 13

thats the formula,
see if u can apply it, and interpret wat the formula means.

Answer 14

f(2)-f(10)/2-10

Answer 15

f(2)=10*2^2
f(2)=10*4
f(2)= 40?

Answer 16

f(10)=10*2^10
f(10)=10*1024
f(10=10240?

Answer 17

yup ! keep going

Answer 18

40-10240/2-10

Answer 19

-10200/-8

Answer 20

1275?

Answer 21

yes, wat units ?

Answer 22

1275 frogs per .... ?

Answer 23

1275 frogs per year?

Answer 24

in ur previous question
f(x) = A*B^x,

x is in years is it ?

Answer 25

yes

Answer 26

then yes, its 1275 frogs per year.

Answer 27

now explain in ur own words, wat the average rate 1275 represents.
use 2-3 sentences

Answer 28

But I am supposed to be looking between the years 3 and 5 and years 5 and 7...

Answer 29

yes it is defenitely an increase,
but why did u calculate the average rate between x = 2 and x = 10 ?
the question is asking u to calculate between x=3 and x=5 right ?

Answer 30

yes.... lol why did we do average rate between x = 2 and x = 10 :o

Answer 31

idk :'(

Answer 32

And also x=5 and x=7

Answer 33

its okay, compute the average rate again between 3 & 5

and, 5 and 7

more practice ! good for u :D

Answer 34

f(b)-f(a)/b-a
f(3)-f(5)/3-5

f(3)=?

Answer 35

Are we still using 10 and 2?

Answer 36

we're done wid 10 and 2 right.
now you're doing average rate between 3 & 5

Answer 37

What are A and B when x=3 ?

Answer 38

\(f(x) = 10*2^x\)

Answer 39

so f(3)=10*2^3
f(3)= 10*8
f(3) = 80

Answer 40

Yes !

Answer 41

f(5)=10*2^5
f(5)=10*32
f(5)=320

Answer 42

yes ! keep going..

Answer 43

80-320/2-10
240/-2
-120

Answer 44

I'm confused :/

Answer 45

80-320/2-10
-240/-2
120

Answer 46

Right, okay. But what I don't understand is the equation is f(b)-f(a)/b-a? how can f(b)=3 and f(a)=5 but a=10 and b =2?

Answer 47

forget about A and B here.

your function is simply : \(f(x) = 10*2^x\)

Answer 48

Ok

Answer 49

average rate of change between \(x = m\) and \(x = n \) is :-

\(\huge \frac{f(m) - f(n)}{m-n}\)

Answer 50

if it helps,
try to use above formula. i thinkg a and b are confusing us

Answer 51

what are m and n

Answer 52

\(f(x) = 10*2^x\)

average rate of change between \(x = 3\) and \(x = 5\) is :-

\(\huge \frac{f(3) - f(5)}{3-5}\)

Answer 53

-240/-2
120

Answer 54

120 frogs per year

Answer 55

yes !

Answer 56

Ok so now 5 and 7

Answer 57

f(5)-f(7)/5-7
f(5)-f(7)/-2

Answer 58

f(5)=10*2^5
f(5)=10*32
f(5) = 320

f(7)=10*2^7
f(7)=10*128
f(7)=1280

Answer 59

320-1280/-2
-960/-2
480

Answer 60

480 frogs per year?

Answer 61

How come it is different? :S

Answer 62

good question,
may be u can try adn tell me why the change is not same :)

Answer 63

hint : our function is not a straight line, to have FIXED slope

Answer 64

Oh because as the time increases so does the populace, directly proportional?

Answer 65

nope. u doing calculus right. use that knowledge