Question

function f, graphed below, is not an invertible function.
to which intervals could we restrict the domain of f to make it an invertible function?
choose all answers that apply:

Answer:

To determine the intervals where we can restrict the domain of \( f \) to make it invertible, we use the Horizontal Line Test: a function is invertible (has an inverse) if and only if no horizontal line intersects its graph more than once (i.e., it is one - to - one).

Step 1: Analyze the graph’s behavior (increasing/decreasing intervals)

The graph of \( f \) is a wave - like curve. To find one - to - one intervals, we identify where the function is strictly increasing or strictly decreasing (since on these intervals, a horizontal line will intersect the graph at most once).

Looking at the graph:

• Intervals where \( f \) is strictly increasing or strictly decreasing are separated by its local maxima and minima. The critical points (local max/min) seem to occur around \( x=-6, - 4, - 2, 2, 4, 6 \) (approximate from the grid).

Step 2: Identify valid intervals for one - to - one behavior

A function is one - to - one on an interval if it is either strictly increasing or strictly decreasing on that interval. Let's consider some possible intervals:

Interval 1: \([-6, - 2]\)

On \([-6, - 2]\), the function first increases to a local maximum at \( x = - 6\) (wait, actually, looking at the graph: from \( x=-6\) to \( x=-2\), the function goes from the left - most point, rises to a peak at \( x=-6\)? No, correction: Let's re - examine. The left - most part: from \( x=-8\) (approx) to \( x=-6\) (local max), then decreases to \( x=-2\) (local min). Wait, maybe better to look at the symmetry. The graph is symmetric? Wait, the key is to find an interval where the function is monotonic (only increasing or only decreasing).

Another approach: The function has “hills” and “valleys”. A function is invertible on an interval if it is monotonic (strictly increasing or strictly decreasing) on that interval.

Let's take the interval \([-2, 2]\): Wait, no. Wait, looking at the graph, if we take an interval like \([-6, - 2]\): From \( x=-6\) (a local maximum) to \( x=-2\) (a local minimum), the function is decreasing. So it is strictly decreasing on \([-6, - 2]\), so it will pass the horizontal line test.

Another interval: \([2, 6]\): From \( x = 2\) (a local maximum) to \( x=6\) (the right - most point), the function first decreases to a local minimum at \( x = 4\) and then increases? No, wait, from \( x=2\) to \( x = 4\), it decreases, and from \( x=4\) to \( x=6\), it increases. So \([2, 6]\) is not monotonic. Wait, maybe \([-2, 2]\): From \( x=-2\) (local min) to \( x = 2\) (local max), the function is increasing. So on \([-2, 2]\), the function is strictly increasing, so it is one - to - one.

Or \([4, 6]\): From \( x = 4\) (local min) to \( x=6\), the function is increasing. So \([4, 6]\) is a strictly increasing interval.

Or \([-6, - 4]\): From \( x=-6\) (local max) to \( x=-4\) (a point), wait, no. Let's think of the standard way: A function \( y = f(x)\) is invertible on an interval \( I\) if \( f\) is one - to - one on \( I\), i.e., if \( f(a)
eq f(b)\) for \( a
eq b\) in \( I\).

Common valid intervals (depending on the graph’s actual shape, but from the given grid):

• \([-6, - 2]\): The function is decreasing (from a local max to a local min), so one - to - one.
• \([2, 6]\): Wait, no, from \( x = 2\) (local max) to \( x=4\) (local min) it decreases, then from \( x=4\) to \( x=6\) it increases. So \([2, 6]\) is not monotonic. Wait, maybe \([-2, 2]\): From \( x=-2\) (local min) to \( x = 2\) (local max), the function is increasing, so one - to - one.
• \([4, 6]\): From \( x = 4\) (local min) to \( x=6\), the function is increasing, so one - to - one.
• \([-6, - 2]\): Decreasing, one - to - one.

Example of correct intervals (assuming the graph’s key points)

Suppose the options include intervals like:

• \(\boldsymbol{[-6, - 2]}\): The function is strictly decreasing here, so it is one - to - one.
• \(\boldsymbol{[2, 6]}\): Wait, no, correction. Wait, from \( x = 2\) (local max) to \( x=4\) (local min) it decreases, then from \( x=4\) to \( x=6\) it increases. So \([2, 6]\) is not monotonic. Wait, maybe \([-2, 2]\): From \( x=-2\) (local min) to \( x = 2\) (local max), the function is increasing, so one - to - one.
• \(\boldsymbol{[4, 6]}\): From \( x = 4\) (local min) to \( x=6\), the function is increasing, so one - to - one.

(Note: Since the actual options are not provided, but the process is: find intervals where the function is strictly increasing or strictly decreasing. For example, if the options are A. \([-6, - 2]\), B. \([-2, 2]\), C. \([2, 6]\), D. \([4, 6]\), then the correct ones are A, B, D (depending on the exact graph). But based on the graph’s shape, intervals where the function is monotonic (only increasing or only decreasing) are valid. )

If we assume the options are, for example:

• Option 1: \([-6, - 2]\) (decreasing, one - to - one)
• Option 2: \([-2, 2]\) (increasing, one - to - one)
• Option 3: \([2, 6]\) (not monotonic, since it decreases then increases)
• Option 4: \([4, 6]\) (increasing, one - to - one)

Then the correct answers would be the intervals where the function is strictly monotonic (increasing or decreasing).

(If you provide the actual options, we can give a more precise answer. But the key is to identify intervals of strict monotonicity.)