Problem 18 Show that the polynomial \(P(x)=... [FREE SOLUTION]

Chapter 7: Problem 18

Show that the polynomial \(P(x)=x^{7}+x^{5}-2 x^{4}+x^{3}-3 x^{2}+7 x-5\) cannot have a negative real root.

Short Answer

Expert verified

The first derivative of the given polynomial \(P(x)\) is \(P'(x) = 7x^6 + 5x^4 - 8x^3 + 3x^2 - 6x + 7\). For negative values of \(x\), the signs of the terms in both \(P(x)\) and \(P'(x)\) alternate, with the leading term of \(P(x)\) being negative and the leading term of \(P'(x)\) being positive. Since all terms in \(P'(x)\) are positive for negative values of \(x\), it is not possible for \(P'(x)\) to have a negative real root. Therefore, the polynomial \(P(x)\) cannot have a negative real root.

Step by step solution

Find the first derivative of P(x)

Find the first derivative of \(P(x)\) by differentiating each term with respect to \(x\). \(P'(x) = 7x^6 + 5x^4 - 8x^3 + 3x^2 - 6x + 7\) Now we'll analyze the behavior of both \(P(x)\) and \(P'(x)\) for negative values of \(x\).

Analyze the signs of P(x) and P'(x) for negative x-values

For negative values of \(x\), the signs of the terms in both \(P(x)\) and \(P'(x)\) alternate. However, all of the terms in \(P'(x)\) will be positive due to the even powers of \(x\). Also, the leading term of \(P(x)\) is \(x^7\), which is negative for negative values of \(x\), and the leading term of \(P'(x)\) is \(7x^6\), which is positive for negative values of \(x\). Given that the signs alternate and the leading terms have different signs, we can now analyze whether \(P(x)\) can have a negative real root.

Check for the possibility of a negative real root

Since the leading term of \(P(x)\) is negative for negative values of \(x\) and the leading term of \(P'(x)\) is positive for negative values of \(x\), if \(P(x)\) has a negative real root, then \(P'(x)\) must have a negative real root as well so that the slope changes from positive to negative. However, as we have already established, all terms in \(P'(x)\) are positive for negative values of \(x\); thus, it's not possible for \(P'(x)\) to have a negative real root. Therefore, \(P(x)\) cannot have a negative real root. Hence, the polynomial \(P(x) = x^7 + x^5 - 2x^4 + x^3 - 3x^2 + 7x - 5\) cannot have a negative real root.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation

In mathematics, differentiation is a technique that helps us find the rate at which a function is changing at any point. When we talk about polynomials, differentiation allows us to calculate the first derivative, which tells us the slope of the polynomial's graph at a given point.

For the polynomial \(P(x) = x^{7}+x^{5}-2x^{4}+x^{3}-3x^{2}+7x-5\), finding its first derivative involves applying differentiation rules to each term separately following these steps:

The derivative of \(x^n\) is \(nx^{n-1}\).
Applying this rule to each term, we get: \(7x^6 + 5x^4 - 8x^3 + 3x^2 - 6x + 7\).
This new polynomial, \(P'(x)\), represents the change in \(P(x)\) with respect to \(x\).

Having the derivative is essential because it helps us understand how the polynomial behaves and predict where it might have roots or changes in direction.

Polynomial Behavior

Understanding polynomial behavior is crucial for predicting how a polynomial functions. This involves analyzing both the polynomial itself and its derivatives to see how it acts over the range of \(x\) values.

Consider the polynomial \(P(x)\) and its first derivative \(P'(x)\). Here's how each behaves for negative \(x\) values:

For \(P(x)\), note that terms with odd powers of \(x\) keep their sign when \(x\) is negative (negative values for negative \(x\)), whereas even powers are always positive.
The leading term \(x^7\) changes sign and becomes negative when \(x\) is negative, indicating \(P(x)\) decreases.
On the other hand, \(P'(x)\) predominantly has even powers, ensuring it stays positive for negative \(x\).

This analysis helps to understand whether decreasing or increasing trends are seen when 'x' changes from negative to positive.

Analysis of Roots

Analyzing roots involves understanding where the polynomial might cross the x-axis. For this polynomial, we focus on whether it can have a negative real root, meaning a value of \(x\) where \(P(x) = 0\) when \(x\) is negative.

We use the derivative \(P'(x)\) to assist in this analysis, as it provides insight into the slope or tangent's direction:

Any potential negative root in \(P(x)\) would require \(P'(x)\) to also indicate a slope change鈥攁 transition from positive to negative.
Given that all terms in \(P'(x)\) are positive for negative \(x\), there is no sign change. This means \(P'(x)\) never decreases or becomes zero for negative \(x\).
Without such a sign reversal in \(P'(x)\), \(P(x)\) itself cannot have a negative real root.

Thus, we conclude there is no negative real root for \(P(x)\), which makes sense given its behavior and derivative characteristics.

91影视

Show that the polynomial \(P(x)=x^{7}+x^{5}-2 x^{4}+x^{3}-3 x^{2}+7 x-5\) cannot have a negative real root.

Short Answer

Step by step solution

Find the first derivative of P(x)

Analyze the signs of P(x) and P'(x) for negative x-values

Check for the possibility of a negative real root

Key Concepts

Differentiation

Polynomial Behavior

Analysis of Roots

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