- \((x - 4)\) is not a factor because \(f(4) = -136
eq 0\).
- There is only one real solution, \(x = -4\), since the factored form is \(-(x^2+1)(x+4)=0\).
- \(f(1) = -10\) is true by direct substitution.
- As \(x \to \infty\), \(f(x) \to -\infty\) is true because the leading term \(-x^3\) dominates.
- The true statements are: \(f(1) = -10\) and as \(x \to \infty\), \(f(x) \to -\infty\).
Explanation
- Problem Analysis
We are given the polynomial function \(f(x) = -x^3 - 4x^2 - x - 4\). To analyze the given statements, we will evaluate each one systematically.
- Is \((x - 4)\) a factor of \(f(x)\)?
- Are there exactly 3 real solutions for \(f(x) = 0\)?
- Does \(f(1) = -10\)?
- Does \(f(x) \to -\infty\) as \(x \to \infty\)?
- Checking if (x-4) is a factor
To check if \((x - 4)\) is a factor, we apply the Factor Theorem, which states that \((x - c)\) is a factor of \(f(x)\) if and only if \(f(c) = 0\).
Substituting \(x = 4\) into \(f(x)\):
\(f(4) = -(4)^3 - 4(4)^2 - (4) - 4\)
\(f(4) = -64 - 4(16) - 4 - 4\)
\(f(4) = -64 - 64 - 4 - 4 = -136\)
Since \(f(4)
eq 0\), \((x - 4)\) is not a factor of \(f(x)\).
- Solving for real roots
To find the real roots, we set \(f(x) = 0\):
\(-x^3 - 4x^2 - x - 4 = 0\)
We can use grouping to factor the expression:
\((-x^3 - 4x^2) - (x + 4) = 0\)
\(-x^2(x + 4) - 1(x + 4) = 0\)
\(-(x^2 + 1)(x + 4) = 0\)
The roots are \(x = -4\) and the roots of \(x^2 + 1 = 0\). The roots of \(x^2 + 1 = 0\) are \(x = i\) and \(x = -i\), which are imaginary.
Thus, there is only one real root, \(x = -4\). The claim that there are exactly 3 real solutions is false.
- Evaluating f(1)
To check if \(f(1) = -10\), we substitute \(x = 1\) into the expression for \(f(x)\):
\(f(1) = -(1)^3 - 4(1)^2 - (1) - 4\)
\(f(1) = -1 - 4(1) - 1 - 4\)
\(f(1) = -1 - 4 - 1 - 4 = -10\)
This statement is true.
- Determining end behavior
The end behavior of a polynomial as \(x \to \infty\) is determined by its leading term, which in this case is \(-x^3\).
As \(x\) becomes very large and positive, \(x^3\) also becomes very large and positive. Therefore, \(-x^3\) becomes very large and negative, approaching \(-\infty\).
Thus, as \(x \to \infty\), \(f(x) \to -\infty\). This statement is true.
- Conclusion
Based on the analysis:
- \((x - 4)\) is a factor: False.
- Exactly 3 real solutions: False.
- \(f(1) = -10\): True.
- As \(x \to \infty\), \(f(x) \to -\infty\): True.
The true statements are \(f(1) = -10\) and as \(x \to \infty\), \(f(x) \to -\infty\).
Examples
Polynomial end behavior is critical in fields like engineering and economics. For example, if you are modeling the profit of a company as a function of the number of items sold, you might find a cubic function where the leading term is negative. This indicates that as production volume increases beyond a certain point, costs (like infrastructure or maintenance) could cause total profit to drop rapidly toward negative infinity, helping business owners determine an optimal production range.