Question

D%200%20%20to%20x%20%3D%201%20and%20Section%20B%20is%20from%20x%20%3D%202%20to%20x%20%3D%203%20.%5C%5CHow%20many%20times%20greater%20is%20the%20average%20rate%20of%20change%20of%20Section%20B%20than%20Section%20A%3F%5C%5C%20Explain%20why%20one%20rate%20of%20change%20is%20greater%20than%20the%20other.%20%5C%5C" id="TexFormula1" title="Given the function f(x) = 5^{x} , Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3 .\\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. \\" alt="Given the function f(x) = 5^{x} , Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3 .\\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. \\" align="absmiddle" class="latex-formula">

Answer 1

Answer:

The average rate of change in section 2 is 25 times larger than the average rate of change in section 1.

Step-by-step explanation:

Recall that the average rate of change of a function $f(x)$ in an interval [a,b] (section of the number line) is defined as:

Rate of change = $\frac{f(b)-f(a)}{b-a}$

Therefore:

1) First interval: evaluating the rate of change from x=0, to x=1 (interval [0, 1]) it becomes

Rate of change = $\frac{f(b)-f(a)}{b-a}=\frac{f(1)-f(0)}{1-0}=\frac{5^1-5^0}{1}=5-1=4$

2) Second interval: we now evaluate the rate of change from x=2 to x=3 (interval [2, 3]), so it becomes

$\frac{f(b)-f(a)}{b-a}=\frac{f(3)-f(2)}{3-2}=\frac{5^3-5^2}{1}=125-25=100$

Therefore, the rate of change in the second interval is much larger than the rate of change in the first one. The second rate of change is in fact 100/4 = 25 times larger than the first rate of change. this is due to the fact that the function is  an exponential function and not a linear function (where the rate of change is constant)