Given the function g(x) = 6(4)x, Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3.

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)


Given the function gx 64x Section A is from x 0 to x 1 and Section B is from x 2 to x 3 Part A Find the average rate of change of each section 4 points Part B H class=

Respuesta :

[tex]\text{A function is a relationship or expression that involves}[/tex] [tex]\text{one or more variables.}[/tex]

[tex]\text{When we are given Composed Functions), we have the ability}[/tex] [tex]\text{to combine them in such a way that the}[/tex] [tex]\text{outcomes of one function becomes the other}[/tex]

[tex]\text{For example - }[/tex]

If [tex]f(x)=3x-1[/tex] and [tex]g(x)=x^3+2[/tex], then what is[tex]f(g(3))[/tex]?

 [tex]g(x)=x^3+2[/tex] 

 [tex]g(x)=(3)^3+2[/tex]

[tex]= 29 [/tex]

[tex]\text{Since}[/tex] [tex]g(3) = 29[/tex] then [tex]f(g(3))=f(29)[/tex]

[tex]\text{Now let's evaluate}[/tex] [tex]f=(29)[/tex]

[tex]f(x) = 3x-1[/tex]

[tex]f(29)=3(29)-1[/tex] 

[tex]=86[/tex]

[tex]f(g(3))=f(29)=86[/tex]

[tex]\text{To find the value of g(x) we need to substitute}[/tex] [tex]\text{a number into the function's formula: }[/tex]

[tex]g\left(x\right)=6\left(4\right)x[/tex] 

[tex] \dfrac{d}{dx} (6*4x)[/tex] 

[tex]6*4\dfrac{d}{dx}(x)[/tex] 

[tex]6*4*1[/tex] 

[tex]= 24[/tex]

[tex]\text{The "rate of change" is the slope of a function .}[/tex]

[tex]\text{Formula:}[/tex]   [tex] \dfrac{\text{(change in f(x) }}{\text{(change in 'x')}} [/tex]

[tex]\text{In Section A:}[/tex]

[tex]\text{Length of section}[/tex] = (1 - 0) = 1

[tex]f(1) = 6(4)x[/tex]

[tex]f(0) = 0[/tex]

[tex]\text{Change in the value of the given function}[/tex] = [tex](24 - 0) = 24[/tex]

 [tex] \dfrac{\text{ (change in the value of the function)}}{\text{(size of the section)}} [/tex] [tex] \dfrac{24}{1} = 24 [/tex]