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Mathematics 8 Online
OpenStudy (anonymous):

Given the function \(f(x) = 5^x\), Section A goes from \(x = 0\) to \(x = 1\). Section B goes from \(x = 2\) to \(x = 3\). Part A: Find the average rate of change for each section. Part B: How many times greater is the average rate of change of Section B than Section A? Explain why one rate of change is greater than the other.

OpenStudy (excalibur0126):

What's the question?

OpenStudy (anonymous):

I'm not too sure how to do this @Excalibur0126

Nnesha (nnesha):

formula to find average rate of change (slope ) \[\huge\rm \frac{ y_2 -y_1 }{ x_2 -x_1 }\] given: x = 0 and x=1 substitute x for 0 and for 1 into \[\rm f(x)=5^x\]to find y `f(x) is same as y

OpenStudy (anonymous):

0 is x^1? what is x^2 and y^2? @Nnesha

Nnesha (nnesha):

it's not x^2 (not squared) \[\huge\rm y_2\] means 2nd y you need two order pairs to find slope (x , y)(x,y) (0 , ?) ( 1,y) you should find y value when x = 0 so replace x with 0 \[\huge\rm f(0)=5^{0}\] f(0)= ??

Nnesha (nnesha):

exponent rule anything to the zero power is one \[\rm (anything)^0 = 1 ~~~(x)^0=1 \]

OpenStudy (anonymous):

So what are the steps to doing something like this in the simplest terms? @Nnesha

Nnesha (nnesha):

well first step) find y when x=0 , x=1 2nd step) find y when x=2 and x=3 when you get all y values then you can apply slope formula to find average rate of change

Nnesha (nnesha):

\[\huge\rm f(0)=5^0 ~~~~~~f(1)=5^1\] like i said f(x) is same as y so you replace f(0) and f(1) with y \[\large\rm y=5^0 ~~~~~~~~~~y=5^1\] simplify that^^ (0 , ?)( 1,?) when x=0 , y = ?? when x=1 y =what ? write your answer where i put the question marks

Nnesha (nnesha):

does it make sence ?

Nnesha (nnesha):

sense *

OpenStudy (anonymous):

I have no idea whats hapenning @Nnesha

Nnesha (nnesha):

5^0 = ?

Nnesha (nnesha):

5 to the zero power equal to what ?

OpenStudy (anonymous):

5?

Nnesha (nnesha):

no did you read my comments ^^??

OpenStudy (anonymous):

*1 (im kind of slow with math)

Nnesha (nnesha):

yes right so when x=0 y = 1 (0 ,1) now replace x with 1 \[y=5^1\] 5^1 is same as 5 right so when x =1 y=5 (0 , 1)(1,5) two order pairs now you can use slope formula to find average rate of change

OpenStudy (anonymous):

thats y2 - y1 over x2 - x1?

Nnesha (nnesha):

yes right use these two points (0,1)(1,5)

OpenStudy (anonymous):

5-1 over 1-0?

OpenStudy (anonymous):

4 over 1 is the slope?

Nnesha (nnesha):

perfect! that's average rate of change when x=0 and x=1 now

Nnesha (nnesha):

now find y when x=2 and x=3

Nnesha (nnesha):

section B^

OpenStudy (anonymous):

(x=2 y=25) (x=3 y=125)

Nnesha (nnesha):

great now find slope!

OpenStudy (anonymous):

100 over 1?

Nnesha (nnesha):

nice now you can write answer for PART B

OpenStudy (anonymous):

Wait what was the answer for part A?

Nnesha (nnesha):

find average rate of change of `each` section average rate of change = `slope` we just found the answer

OpenStudy (anonymous):

Its just the two slopes?

Nnesha (nnesha):

yes right!

OpenStudy (anonymous):

oh okay so For A should I write 4 over 1 and 100 over 1?

Nnesha (nnesha):

2/1 is same as 2 so 4/1 is same as 4

OpenStudy (anonymous):

so i write 4 and 100?

Nnesha (nnesha):

ye

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