Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

Solve for \(x\), expressing your answer in interval notation. (a) \((x+1)\left(x^{2}+2 x-7\right) \geq x^{2}-1\) (b) \(x^{4}-2 x^{2} \geq 8\) (c) \(\left(x^{2}+1\right)^{2}-7\left(x^{2}+1\right)+10<0\)

Step-by-Step Solution

Answer

(a) \([1, \infty)\) including \(1\); (b) \([-\infty, -2] \cup [2, \infty)\); (c) \((-2, -1) \cup (1, 2)\).

Step 1: Simplify Inequality 1

Start by simplifying the inequality \((x+1)(x^2 + 2x - 7) \geq x^2 - 1\). Expand the left side and move all terms to one side to form a single inequality: \(x^3 + 2x^2 - 7x + x^2 + 2x - 7 \geq x^2 - 1\). This simplifies to \(x^3 + 3x^2 - 5x - 6 \geq x^2 - 1\). Subtract \(x^2 - 1\) from both sides to get \(x^3 + 2x^2 - 5x - 5 \geq 0\).

Step 2: Factor and Analyze Sign Changes

Factor the expression \(x^3 + 2x^2 - 5x - 5\) by finding rational roots or using synthetic division. Let's assume a root is found at \(x = 1\), leaving us with \((x-1)(x^2 + 3x + 5)\). Use the quadratic formula to check that \(x^2 + 3x + 5\) has no real roots, since the discriminant \(3^2 - 4 \times 5 < 0\). The sign change happens only at \(x = 1\), indicating intervals \((x - 1 \geq 0)\).

Step 3: Solve Inequality 2

For the second inequality \(x^4 - 2x^2 \geq 8\), rewrite it as \(x^4 - 2x^2 - 8 \geq 0\). Let \(y = x^2\), which transforms the inequality to \(y^2 - 2y - 8 \geq 0\). Factor it as \((y-4)(y+2) \geq 0\). Analyze the intervals where \(y = x^2\) satisfies \(y \geq 4\) or \(y \leq -2\). Only \(y \geq 4\) is possible because \(x^2\) is non-negative. Solve \(x^2 \geq 4\), giving \(x \leq -2\) or \(x \geq 2\).

Step 4: Simplify Inequality 3

Convert the expression \((x^2 + 1)^2 - 7(x^2 + 1) + 10 < 0\) by substituting \(z = x^2 + 1\). The inequality becomes \(z^2 - 7z + 10 < 0\). Factor as \((z-2)(z-5) < 0\). Analyze the roots with intervals to find \(2 < z < 5\).

Step 5: Interpret Final Range for Inequality 3

Translate back from \(2 < z < 5\) into terms of \(x^2\): \(2 < x^2 + 1 < 5\), simplifying to \(1 < x^2 < 4\). Thus, \(x^2\) in the range implies \(-2 < x < -1\) or \(1 < x < 2\), excluding endpoints for strict inequality.

Key Concepts

Polynomial Inequalities

Polynomial inequalities are expressions involving a polynomial that lead to an inequality rather than an equality. Solving these often involves finding where the polynomial is positive or negative.
For example, consider the inequality \(f(x) \geq 0\). The solution involves determining for which values of \(x\) the polynomial \(f(x)\) is above or equal to zero. To solve polynomial inequalities, you generally:

• Expand and simplify the polynomial.
• Move all terms to one side of the inequality to form \(f(x) \geq 0\) or \(f(x) \leq 0\).
• Factor the polynomial, if possible, or find its roots using methods such as synthetic division or the rational root theorem.
• Analyze the sign changes around the roots to determine the intervals where the inequality holds.

In our example, we expanded and simplified the given polynomial, then analyzed the sign changes at the roots to determine the solution interval.

Rational Roots

Rational roots form an essential part of solving polynomial inequalities as they provide the points where the polynomial changes sign. The Rational Root Theorem helps in identifying potential rational roots of a polynomial by examining the factors of the constant term and the leading coefficient.
To apply the Rational Root Theorem:

• Identify the possible rational roots as \(\frac{p}{q}\), where \(p\) is a factor of the constant term, and \(q\) is a factor of the leading coefficient.
• Test these values by substituting them back into the polynomial to identify any actual roots.
• When a rational root is identified, it can be used to factor the polynomial further, simplifying the inequality solving process.

In our solution, finding a root at \(x = 1\) allowed us to factor the polynomial, focusing on intervals around this root to check where the polynomial is positive or negative.

Quadratic Inequalities

Quadratic inequalities involve expressions of the form \(ax^2 + bx + c > 0\) or \(ax^2 + bx + c < 0\). Solving these requires looking for values of \(x\) where the quadratic expression changes sign.
To solve quadratic inequalities:

• Rewrite the inequality as a standard quadratic: \(ax^2 + bx + c > 0\) or \(ax^2 + bx + c < 0\).
• Factor the quadratic, if possible, which transforms it into a product of expressions like \((x-r_1)(x-r_2)\).
• If factoring is not possible, use the quadratic formula to determine the roots of the quadratic.
• Analyze the intervals defined by these roots by selecting test points within each interval to see where the quadratic expression holds the desired inequality.
• Summarize the solution in terms of intervals that satisfy the inequality.

Converting the given polynomial using the substitution technique and analyzing the resulting inequality allowed us to determine the suitable range for \(x\) that meets the inequality conditions.