- A function is invertible if it is a bijection, meaning it must be injective (one-to-one).
- By checking the table, we observe that \(g(-1) = 4\) and \(g(1) = 4\).
- Since two distinct inputs result in the same output, the function is not injective.
- Consequently, the function \(g\) is \(\boxed{\text{not invertible}}\).
Explanation
- Understanding Invertibility and Injectivity
To determine if the function \(g\) is invertible, we must evaluate whether it satisfies the condition of being a bijection. For a function defined on a discrete set of points, this requires the function to be injective (one-to-one). A function is injective if every distinct input \(x\) in the domain maps to a unique output \(g(x)\). If any two distinct inputs \(x_1\) and \(x_2\) result in the same output, meaning \(g(x_1) = g(x_2)\), then the function is not injective and, consequently, not invertible.
- Analyzing the Function Values
We analyze the provided table of values for the function \(g(x)\):
- At \(x = -3, g(x) = 11\)
- At \(x = -2, g(x) = 6\)
- At \(x = -1, g(x) = 4\)
- At \(x = 0, g(x) = 3\)
- At \(x = 1, g(x) = 4\)
By inspecting these values, we look for any instances where the same output value appears more than once for different inputs. We observe that for both \(x = -1\) and \(x = 1\), the function produces the same output value: \(g(-1) = 4\) and \(g(1) = 4\).
- Conclusion on Invertibility
Since we have found that two distinct domain values, \(x_1 = -1\) and \(x_2 = 1\), map to the same range value \(g(x_1) = g(x_2) = 4\), the function \(g\) fails the horizontal line test. Because it is not injective, it cannot be a bijection. Therefore, the function \(g\) is not invertible. If we attempted to invert the function, the value \(4\) would map to both \(-1\) and \(1\), which violates the fundamental definition of a function. Thus, the conclusion is that the function \(g\) is not invertible.
Examples
Think of invertibility like a locker system where each key (the input \(x\)) should open exactly one specific locker (the output \(g(x)\)). If two different keys open the same locker, you cannot uniquely determine which key was used just by looking at the opened locker. In this problem, because the inputs \(-1\) and \(1\) both "open the same locker" (they both map to the value \(4\)), you cannot reverse the process to identify the original input, meaning the function is not invertible. This concept is crucial in fields like cryptography, where each piece of encrypted data (the output) must be linked back to a unique original message (the input).