91Ó°ÊÓ

Real zeros of a polynomial function are the values of x for which the function evaluates to zero. To find these, we look at the points where the graph of the function intersects the x-axis. In our exercise, we need to determine that there are no real zeros for the polynomial function, \(f(x) = x^5 - 3x^3 + x + 2\), when \(x < -3\). This involves analyzing where \(f(x) = 0\). To start, we substitute and evaluate the polynomial at \(x = -3\), which gives us \(f(-3) = -163\). Since \(-163 < 0\), it indicates that at \(x = -3\), the function value is negative and does not intersect the x-axis, suggesting no zero at \(x=-3\). To further understand this, we need to analyze the behavior of the function and its derivative in this region.

Derivative analysis helps us understand how a function behaves, such as whether it is increasing or decreasing. The first derivative, \(f'(x)\), indicates the slope of the function at any given point. For \(f(x) = x^5 - 3x^3 + x + 2\), the first derivative is \(f'(x) = 5x^4 - 9x^2 + 1\). By analyzing \(f'(x)\), we can determine whether the function is increasing or decreasing.
For \(x < -3\), all terms of the derivative \(5x^4 - 9x^2\) are positive since the exponents are even. This means \(5x^4\) and \(-9x^2\) are both positive. Therefore, \(5x^4 - 9x^2 + 1\) remains positive for \(x < -3\), indicating that the function \(f(x)\) is always increasing in this interval. This constant increase confirms that the value of \(f(x)\) will not change signs (cross the x-axis) for \(x < -3\), part of reason why we know there are no zeros in that interval.

Evaluating a polynomial at specific points helps us understand the function's behavior at those points. The process involves substituting a particular value of \(x\) into the polynomial and simplifying to find \(f(x)\).
In our exercise, we substitute \(x = -3\) into \(f(x) = x^5 - 3x^3 + x + 2\). Evaluating step-by-step:

• \((-3)^5 = -243\)
• \(-3(-3)^3 = -3(-27) = 81\)
• Combine the results: \(-243 + 81 - 3 + 2 = -163\)

Thus, \(f(-3) = -163\). Because \(-163\) is a negative value, and the function is increasing for \(x < -3\), it confirms that the value of \(f(x)\) does not cross zero (the x-axis), which helps establish there are no real zeros for \(x < -3\).

To understand the behavior of a polynomial function, we look at both its evaluation and derivative. The polynomial function \(f(x) = x^5 - 3x^3 + x + 2\) is non-linear and consists of different degrees of \(x\). Its behavior changes depending on the values of \(x\).
By evaluating \(f(-3) = -163\), we see that the function is negative at \(x = -3\). To further understand how the function behaves around \(x = -3\), we use its derivative: \(f'(x) = 5x^4 - 9x^2 + 1\). Since \(f'(x)\) is positive for \(x < -3\), the function \(f(x)\) is increasing in this interval. This means that as \(x\) becomes more negative, \(f(x)\) gets smaller, confirming \(f(x)\) is below the x-axis and does not have any real zero less than \(-3\). This comprehensive approach shows how both polynomial evaluation and derivative analysis together help us determine the behavior and characteristics of the function, confirming no real zeros exist for \(x < -3\).