Question

This table shows some values of the function f(x)=
4
3
x–2.
x
–
1 0 1 2 3
f(x)
–
10
3
–
2
–
2
3
2
3
2
Complete the sentences.
When the x-value changes by 1 unit, the function's value changes by
. So, f(x) demonstrates that
functions change by equal
over equal intervals.

Answer 1

To complete the sentences, we first need to look at how the function values change as the x-values increase by 1 unit. Let's analyze the given values:

- When \( x = -1 \), \( f(-1) = -10/3 \)
- When \( x = 0 \), \( f(0) = -2 \)
- When \( x = 1 \), \( f(1) = 2/3 \)
- When \( x = 2 \), \( f(2) = 2 \)
- When \( x = 3 \), \( f(3) = 2 \)

Now, let's calculate the change in the function's value \( f(x) \) for each increase of 1 in \( x \):

- From \( x = -1 \) to \( x = 0 \):
\( f(0) - f(-1) = -2 - (-10/3) = -2 + 10/3 = -6/3 + 10/3 = 4/3 \)

- From \( x = 0 \) to \( x = 1 \):
\( f(1) - f(0) = 2/3 - (-2) = 2/3 + 6/3 = 8/3 \)

- From \( x = 1 \) to \( x = 2 \):
\( f(2) - f(1) = 2 - (2/3) = 6/3 - 2/3 = 4/3 \)

- From \( x = 2 \) to \( x = 3 \):
\( f(3) - f(2) = 2 - 2 = 0 \)

Now, let's summarize the change in f(x):
- between x = -1 to 0: increases by \( \frac{4}{3} \)
- between x = 0 to 1: increases by \( \frac{8}{3} \)
- between x = 1 to 2: increases by \( \frac{4}{3} \)
- between x = 2 to 3: stays the same (0 increase)

Notice that when we say the function's value changes by equal amounts over equal intervals, it applies more to functions that are linear where there's a constant change in \( f(x) \). In this case, you can see differing increases.

So, looking into the completed sentences:
"When the x-value changes by 1 unit, the function's value changes by \( \frac{4}{3} \) on average across some intervals." So, \( f(x) \) demonstrates that linear functions change by equal **amounts** over equal intervals.

To finalize it:
**When the x-value changes by 1 unit, the function's value changes by 4/3 on average. So, f(x) demonstrates that linear functions change by equal amounts over equal intervals.**