MATH SOLVE

4 months ago

Q:
# Given the function g(x) = 4(3)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)Give me a simple answer and you can keep these points.

Accepted Solution

A:

[tex]\bf slope = m = \cfrac{rise}{run} \implies
\cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby
\begin{array}{llll}
average~rate\\
of~change
\end{array}\\\\
-------------------------------[/tex]

[tex]\bf \stackrel{section~A}{gf(x)=4(3)^x \qquad \begin{cases} x_1=1\\ x_2=2 \end{cases}}\implies \cfrac{g(2)-g(1)}{2-1}\implies \cfrac{4(3)^2-4(3)^1}{1} \\\\\\ \cfrac{36-12}{1}\implies \boxed{24} \\\\\\ \stackrel{section~B}{gf(x)=4(3)^x \qquad \begin{cases} x_1=3\\ x_2=4 \end{cases}}\implies \cfrac{g(4)-g(3)}{4-3}\implies \cfrac{4(3)^4-4(3)^3}{1} \\\\\\ \cfrac{324-108}{1}\implies \boxed{216}[/tex]

part B) well, you already know.

why is that? well, because 4(3)ˣ is an exponential function, so the jumps from one point to another are large.

[tex]\bf \stackrel{section~A}{gf(x)=4(3)^x \qquad \begin{cases} x_1=1\\ x_2=2 \end{cases}}\implies \cfrac{g(2)-g(1)}{2-1}\implies \cfrac{4(3)^2-4(3)^1}{1} \\\\\\ \cfrac{36-12}{1}\implies \boxed{24} \\\\\\ \stackrel{section~B}{gf(x)=4(3)^x \qquad \begin{cases} x_1=3\\ x_2=4 \end{cases}}\implies \cfrac{g(4)-g(3)}{4-3}\implies \cfrac{4(3)^4-4(3)^3}{1} \\\\\\ \cfrac{324-108}{1}\implies \boxed{216}[/tex]

part B) well, you already know.

why is that? well, because 4(3)ˣ is an exponential function, so the jumps from one point to another are large.