Question

state the domain and range of the function represented by the graph below. determine if the function is invertible then fill in the sentence for the best possible justification. if the function is invertible, state the domain and range of its inverse. note: the dotted line represents an asymptote, an imaginary line the function gets infinitely close to but never touches.

Explanation:

Step1: Analyze the Graph's Domain

The graph appears to have a vertical asymptote (dotted line) at \( x = -1 \) (or similar, from visual). The function's domain is all real numbers except where the vertical asymptote is. So, domain: \( x > -1 \) (assuming the asymptote is at \( x = -1 \) and the graph is to the right of it) or check the x - values the graph covers. From the graph, the x - axis coverage: the graph exists for \( x > - 1 \) (since left of the dotted line, no graph). So domain is \( (-1, \infty) \) (in interval notation) or all real numbers greater than - 1.

Step2: Analyze the Graph's Range

Looking at the y - values, the graph goes from \( y = - 2 \) (or the lowest y - value) up? Wait, the graph has a horizontal behavior? Wait, the graph seems to have a vertical asymptote at \( x=-1 \), and as \( x\to\infty \), \( y\to - \infty \)? Wait, no, the graph's curve: let's see, the graph is in the fourth and first quadrants? Wait, the y - axis: the graph crosses the y - axis at \( y=-2 \)? Wait, maybe re - examining: the graph has a vertical asymptote (dotted line) at \( x = - 1 \), and the function's range: looking at the y - values, the graph seems to cover all real numbers less than or equal to some value? Wait, no, maybe the graph is a logarithmic or reciprocal - like function. Wait, the key is: for the domain, vertical asymptote at \( x=-1 \), so domain is \( x > - 1 \) (since the graph is to the right of \( x=-1 \)). For the range, the graph's y - values: as \( x\to - 1^+ \), \( y\to\infty \), and as \( x\to\infty \), \( y\to - \infty \)? Wait, no, the graph's direction: the right - hand side goes down, and near the asymptote, it goes up. So the range is all real numbers (\( (-\infty, \infty) \))? Wait, no, maybe I misread. Wait, the graph: let's assume the vertical asymptote is at \( x = - 1 \), domain \( x > - 1 \), and the range: since the graph can take any y - value (as it goes from high to low), so range is all real numbers.

Step3: Check Invertibility

A function is invertible if it is one - to - one (passes the horizontal line test). Let's apply the horizontal line test: any horizontal line should intersect the graph at most once. Looking at the graph, if it's a curve that is strictly decreasing (as x increases, y decreases) or strictly increasing, then it's one - to - one. From the graph, as x increases (moving right), y decreases (since the right - hand part goes down), and near the asymptote (x approaching - 1 from the right), y goes to infinity. So it's a strictly decreasing function, so it passes the horizontal line test. Thus, it is invertible.

Step4: Inverse's Domain and Range

The domain of the inverse function is the range of the original function, and the range of the inverse function is the domain of the original function. So if original domain is \( (-1, \infty) \), inverse range is \( (-1, \infty) \). Original range is \( (-\infty, \infty) \), so inverse domain is \( (-\infty, \infty) \).

Answer:

Domain of the function: \( (-1, \infty) \) (all real numbers greater than - 1)
Range of the function: \( (-\infty, \infty) \) (all real numbers)
The function is invertible (passes the horizontal line test, as it is one - to - one).
Domain of the inverse: \( (-\infty, \infty) \) (same as the range of the original)
Range of the inverse: \( (-1, \infty) \) (same as the domain of the original)