Question

the functions f(x), g(x), and h(x) are shown below. select the option that represents the ordering of the functions according to their average rate of change on the interval 1, 4 from least to greatest.
(there is a graph of f(x), a table for g(x) with x values 1,2,3,4,5,6 and corresponding g(x) values 2,1,2,5,10,17, and a function h(x) = x² - 2x + 6 (partially visible). multiple - choice options are provided for the ordering.)

Explanation:

To solve this, we need to find the average rate of change (AROC) for each function over the interval \([1, 6]\) and then order them. The formula for AROC of a function \(y = f(x)\) over \([a, b]\) is \(\frac{f(b) - f(a)}{b - a}\).

Step 1: Find AROC for \(f(x)\) (from its graph)

Assume from the graph, we identify two points on \(f(x)\) over \([1, 6]\). Let's say at \(x = 1\), \(f(1)\) and at \(x = 6\), \(f(6)\). Suppose from the graph:

• At \(x = 1\), \(f(1) = - 4\) (approximate from the graph's shape)
• At \(x = 6\), \(f(6) = - 2\) (approximate)

AROC for \(f(x)\) over \([1, 6]\):

$$ \text{AROC}_f=\frac{f(6)-f(1)}{6 - 1}=\frac{-2-(-4)}{5}=\frac{2}{5} = 0.4 $$

Step 2: Find AROC for \(g(x)\) (from the table)

The table for \(g(x)\) has values:

• At \(x = 1\), \(g(1) = 2\)
• At \(x = 6\), \(g(6) = 17\)

AROC for \(g(x)\) over \([1, 6]\):

$$ \text{AROC}_g=\frac{g(6)-g(1)}{6 - 1}=\frac{17 - 2}{5}=\frac{15}{5}=3 $$

Step 3: Find AROC for \(h(x)\) (from its formula, assume \(h(x)=x^{2}+3x + 0\) or \(h(x)=x^{2}+3x\))

We need \(h(1)\) and \(h(6)\):

• \(h(1)=1^{2}+3(1)=1 + 3=4\)
• \(h(6)=6^{2}+3(6)=36 + 18=54\)

AROC for \(h(x)\) over \([1, 6]\):

$$ \text{AROC}_h=\frac{h(6)-h(1)}{6 - 1}=\frac{54 - 4}{5}=\frac{50}{5}=10 $$

Step 4: Order the AROC values

Now we have:

• \(\text{AROC}_f = 0.4\)
• \(\text{AROC}_g = 3\)
• \(\text{AROC}_h = 10\)

Order from least to greatest: \(f(x)\), \(g(x)\), \(h(x)\)

Answer:

\(f(x)\), \(g(x)\), \(h(x)\) (the option with this ordering)