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 is a function (let's assume it's an exponential or logarithmic - like curve). The vertical dashed line is at \(x = - 4\) (from the graph's grid, the dashed line is at \(x=-4\)). The graph exists for all \(x\) values greater than \(-4\) (since the curve is to the right of \(x = - 4\)). So the domain of the function \(f(x)\) is \(x>-4\) or in interval notation \((-4,\infty)\).

Step2: Analyze the Graph's Range

Looking at the \(y\) - values the graph covers. The graph goes from \(y = -\infty\) (as \(x\) approaches \(-4\) from the right, the \(y\) - value goes down) up to \(y=\infty\) (as \(x\) increases, the \(y\) - value can go up, and as \(x\) approaches \(\infty\), the \(y\) - value behavior? Wait, no, looking at the graph, the curve starts from the left (near \(x=-4\)) going down? Wait, maybe it's a logarithmic function. Wait, the graph intersects the \(y\) - axis? Wait, the graph's \(y\) - values: let's see the direction. The curve is decreasing from left to right? Wait, no, the arrow on the right end of the curve is going down? Wait, maybe I misread. Wait, the \(x\) - axis is horizontal, \(y\) - axis vertical. The dashed line is \(x=-4\) (vertical asymptote). The graph is to the right of \(x = - 4\). For the range, let's see the \(y\) - values. The graph seems to cover all real numbers? Wait, no, maybe it's a logarithmic function. Wait, if it's a function like \(y=\log(x + 4)\) shifted? No, maybe it's an exponential function reflected. Wait, the key is: for a function, domain is all \(x\) where it's defined, range is all \(y\) it takes.

Wait, the problem also asks if it's invertible. A function is invertible if it's one - to - one (passes the horizontal line test). Let's check the horizontal line test: if any horizontal line intersects the graph more than once, it's not invertible. Looking at the graph, since it's a curve that is either strictly increasing or strictly decreasing, it should pass the horizontal line test. So it is invertible.

Step3: Domain and Range of Inverse Function

For an invertible function \(y = f(x)\), the domain of \(f^{-1}(x)\) is the range of \(f(x)\) and the range of \(f^{-1}(x)\) is the domain of \(f(x)\).

From the graph, the domain of \(f(x)\) is \(x>-4\) (so \((-4,\infty)\)) and the range of \(f(x)\) is all real numbers (\((-\infty,\infty)\)) because the graph extends from \(y = -\infty\) to \(y=\infty\) (assuming the curve's \(y\) - values cover all real numbers as \(x\) varies over \((-4,\infty)\)). So the domain of \(f^{-1}(x)\) is the range of \(f(x)\), which is \((-\infty,\infty)\) and the range of \(f^{-1}(x)\) is the domain of \(f(x)\), which is \((-4,\infty)\).

(Note: The initial analysis of the range might need correction. Let's re - examine the graph. The graph has a vertical asymptote at \(x=-4\). The curve is decreasing (from left to right, the \(y\) - value decreases? Or increasing? Wait, the arrow on the right end of the curve: if the curve is going down as \(x\) increases, then it's a strictly decreasing function. A strictly decreasing function is one - to - one, so it's invertible.

For the domain of \(f\): all \(x\) such that \(x>-4\) (since the vertical asymptote is \(x = - 4\) and the graph is to the right of it).

For the range of \(f\): since the function is strictly decreasing (or increasing) and as \(x\) approaches \(-4^+\) (from the right of \(x=-4\)), the \(y\) - value approaches \(\infty\) (if it's decreasing) or \(-\infty\) (if it's increasing), and as \(x\) approaches \(\infty\), the \(y\) - value approaches \(-\infty\) (if decreasing) or \(\infty\) (if increasing). Wait, maybe the graph is of a function like \(y=\ln(x + 4)\) reflected over the \(x\) - axis? No, \(\ln(x + 4)\) has domain \(x>-4\) and range \((-\infty,\infty)\) and is strictly increasing. If our graph is strictly decreasing, it could be \(y =-\ln(x + 4)\), which also has domain \(x>-4\) and range \((-\infty,\infty)\) and is strictly decreasing (hence one - to - one, invertible).

So domain of \(f\): \(x>-4\) (interval \((-4,\infty)\))

Range of \(f\): all real numbers (interval \((-\infty,\infty)\))

Since \(f\) is invertible (passes horizontal line test, as it's strictly monotonic), domain of \(f^{-1}\) is the range of \(f\): \((-\infty,\infty)\)

Range of \(f^{-1}\) is the domain of \(f\): \((-4,\infty)\)

Answer:

Domain of \(f\): \((-4,\infty)\), Range of \(f\): \((-\infty,\infty)\), The function is invertible (because it passes the horizontal line test, being strictly monotonic), Domain of \(f^{-1}\): \((-\infty,\infty)\), Range of \(f^{-1}\): \((-4,\infty)\)