Verified Answer :

  • Evaluate the \(y\)-intercepts: both \(f(0)\) and \(g(0)\) equal \(-1\).
  • Compare end behaviors: \(f(x)\) behaves differently as \(x \to -\infty\) compared to \(g(x)\).
  • Examine \(x\)-intercepts: the root of \(g(x)\) is not a root of \(f(x)\).
  • Compare ranges: \(f(x)\) has an infinite range, while \(g(x)\) is bounded below by \(-2\).
  • Final Answer: The common feature is the \(y\)-intercept. \(\boxed{y\text{-intercept}}\)

Explanation

  1. Evaluating the y-intercepts

We evaluate the properties of the functions \(f(x) = x^3 + x^2 - 6x - 1\) and \(g(x) = e^x - 2\) to determine which they have in common. We begin by examining the \(y\)-intercept of each function. The \(y\)-intercept occurs when \(x = 0\).
For \(f(x)\):
\[f(0) = (0)^3 + (0)^2 - 6(0) - 1 = -1\]
For \(g(x)\):
\[g(0) = e^0 - 2 = 1 - 2 = -1\]
Since both functions evaluate to \(-1\) at \(x = 0\), they share the same \(y\)-intercept at the point \((0, -1)\).

  1. Analyzing the end behavior

Next, we consider the end behavior.
For the cubic polynomial \(f(x) = x^3 + x^2 - 6x - 1\), as \(x \to \infty\), \(f(x) \to \infty\), and as \(x \to -\infty\), \(f(x) \to -\infty\).
For the exponential function \(g(x) = e^x - 2\), as \(x \to \infty\), \(g(x) \to \infty\), but as \(x \to -\infty\), \(g(x) \to -2\).
Because their limits as \(x \to -\infty\) are different (\(-\infty\) for \(f(x)\) vs \(-2\) for \(g(x)\)), the end behaviors are not in common.

  1. Checking the x-intercepts

Now we analyze the \(x\)-intercepts. For \(g(x) = 0\), we have \(e^x - 2 = 0\), which gives \(e^x = 2\), so \(x = \ln(2) \approx 0.693\).
For \(f(x) = 0\), we have \(x^3 + x^2 - 6x - 1 = 0\). By testing the value \(x = \ln(2) \approx 0.693\) in \(f(x)\), we find \(f(\ln(2)) \approx (0.693)^3 + (0.693)^2 - 6(0.693) - 1 \approx 0.333 + 0.480 - 4.158 - 1 \approx -4.345\), which is not equal to \(0\). Thus, the functions do not share the same \(x\)-intercepts.

  1. Determining the range and concluding the common feature

Finally, we look at the range.
The range of the cubic polynomial \(f(x)\) is all real numbers, \((-\infty, \infty)\).
The range of the exponential function \(g(x) = e^x - 2\) is \((-2, \infty)\) because \(e^x\) is always positive.
Since these ranges are clearly different, this feature is not in common.
Having analyzed all options, we conclude that the only shared feature is the \(y\)-intercept. The final answer is the \(y\)-intercept.

Examples

In engineering, identifying common features of different mathematical models is essential. For instance, if you are modeling two different physical processes—one using a cubic growth function and another using exponential decay—determining if they share common initial states (represented by the \(y\)-intercept at time \(t=0\)) allows you to predict if both systems start from the same baseline value, which is vital for synchronization and calibration in automated control systems.

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