Question

Help! Will medal. -- Open Ended Response

Given the function f(x) = 4(2)x, Section A is from x = 1 to x = 2 and Section B is from x = 3 to x = 4.

Part A: Find the average rate of change of 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.

Answer 1

Is the function f(x) =4*2x?

Answer 2

f(x) = 4(2)^x, sorry.

Answer 3

Start by taking out the first order derivative of f(x)

Answer 4

Average rate of change is basically asking for the slope of the line through the two points on the graph of the function.
\[arc = \frac{ f(b)-f(a) }{ b-a }\]
So for your first section, the average rate of change between x = 1 and x = 2 is
\[arc = \frac{ f(2)-f(1) }{ 2-1 }\]

Answer 5

Sorry, slope is my weak point. So for section b it would be

\[arc = f(4) - f(2) \over 4-1\]
based on the previous equation?

Sorry if this is wrong, I am just trying to understand and solve the question. @peachpi

Answer 6

no problem.
the second second asks for rate of change between 3 and 4, so they'll be the numbers you use.
\[arc = \frac{ f(4)-f(3) }{ 4-3 }\]
The numbers you use for each section have to be given to you. You don't use numbers from previous parts. Also, make sure the x values you use in the to numerator and denominator match.

Answer 7

Oh!! I meant to put those numbers, not the 1 & 2! Ah silly me. Typo.

Answer 8

ok cool.
from there you just plug in the x-values into the function to get a final number

Answer 9

I'm in a school with bright lights staring at a computer for a constant 8 hours, so my eyes get all dry and itchy and the headache starts to form, and from there I lose concentration.

How do I go about solving the function? Wouldn't the answer just essentially be \[f\]

Answer 10

Plug in x-values
For example for section a

\[f(4) = 4(2)^4 = 4(16) = 64\]
\[f(3) = 4(2)^3=4(8)=32\]

So you have

\[arc = \frac{ 64-32 }{ 4-3 }=\frac{32}{1}=32\]

Answer 11

So for section b
\[f(2) = 4(2)^2 = 4(4) = 16\]
\[f(1) = 4(2)^1 = 4(2) = 8\]
So
\[arc = \frac{ 16 - 8 }{ 2 - 1} = \frac{ 8 }{ 1 } = 8\]

Answer 12

yes

Answer 13

THANK YOU SO MUCH FOR YOUR HELP!!

Answer 14

you're welcome