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. (4 points)
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. (6 points)

Step-by-step explanation:

I'm sure you want your functions to appear as perfectly formed as possible so that others can help you.  f(x) = 4(2)x should be written with the " ^ " sign to denote exponentation:  f(x) = 4(2)^x

                                                                                      f(b) - f(a)

The formula for "average rate of change" is a.r.c. =

                                                                                           b - a

                                    change in function value

This is equivalent to  

                                            change in x value

For Section A:  x changes from 1 to 2 and the function changes from 4(2)^1 to  4(2)^2:  8 to 16.  Thus, "change in function value" is 8 for a 1-unit change in x from 1 to 2.  Thus, in this Section, the a.r.c. is:

                 8

                = 8 units    (Section A)

                  1

Section B:  x changes from 3 to 4, a net change of 1 unit:  f(x) changes from

4(2)^3 to 4(2)^4, or 32 to 256, a net change of 224 units.  Thus, the a.r.c. is

        224 units

      = 224 units (Section B)

            1 unit

The a.r.c for Section B is 28 times greater than the a.r.c. for Section A.

This change in outcome is so great because the function f(x) is an exponential function; as x increases in unit steps, the function increases much faster (we say "exponentially").

answer: y=0 if that is what u r looking for

explanation:

infrared radiation has wavelength in the range between 700 nm and 1 mm. lets' convert these values into scientific notation, keeping in mind that:

let's start from the lower value, we have:

(we moved the dot two places to the left, so we increased the exponent by 2 units, from -9 to -7)

the higher value is:

so, the range of wavelength of infrared radiation is: