Question

the function f(x) has an average rate of change of -2 over the interval -1 ≤ x ≤ 1. which graph shows y = f(x) under the graph of function f(x)?

Explanation:

To solve this, we analyze the average rate of change (slope of the secant line) over \(-1 \leq x \leq 1\). The average rate of change formula is \(\frac{f(1) - f(-1)}{1 - (-1)}=\frac{f(1)-f(-1)}{2}\). A rate of change of \(2\) means \(\frac{f(1)-f(-1)}{2}=2\), so \(f(1)-f(-1) = 4\) (i.e., \(f(1)=f(-1)+4\), so \(f(1)>f(-1)\), increasing on \([-1,1]\)).

Analyze each graph:

1. First graph (line with negative slope): Over \(-1 \leq x \leq 1\), \(f(1)<f(-1)\) (decreasing), so rate of change is negative. Eliminate.
2. Second graph (upward parabola, vertex at \(x=0\)): Symmetric about \(y\)-axis. \(f(1)=f(-1)\), so \(f(1)-f(-1)=0\), rate of change \(0\). Eliminate.
3. Third graph (downward parabola, vertex at \(x=0\)): Symmetric about \(y\)-axis. \(f(1)=f(-1)\), so \(f(1)-f(-1)=0\), rate of change \(0\). Eliminate.
4. Fourth graph (line with positive slope): Over \(-1 \leq x \leq 1\), \(f(1)>f(-1)\) (increasing). Calculate the slope: If we assume grid units (e.g., from \(x=-1\) to \(x=1\), the line rises by \(4\) units), slope \(=\frac{4}{2}=2\), matching the rate of change.

Answer:

The fourth graph (the last linear graph with positive slope)