/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Solve. $$4 x^{3}-7 x^{2} \leq ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 39

Solve. $$4 x^{3}-7 x^{2} \leq 15 x$$

Short Answer

Expert verified
The solution is \((-\infty, -\frac{5}{4}] \cup [0, 3]\).

Step by step solution

01

Bring the inequality to standard form

Subtract 15x from both sides of the inequality: $$4x^3 - 7x^2 - 15x \rightarrow 0 $$ The inequality now becomes: \[4x^3 - 7x^2 - 15x \leq 0\] This is the standard form.
02

Factorize the polynomial expression

Factor out the common term x: $$x(4x^2 - 7x - 15) \leq 0$$. Next, factorize the quadratic: $$x(4x + 5)(x - 3) \rightarrow 0$$
03

Determine the critical points

Set each factor to 0 and solve for x: \( x = 0 \), \(4x + 5 = 0 \rightarrow x = -\frac{5}{4} \), and \(x - 3 = 0 \rightarrow x = 3 \).
04

Analyze test intervals

Use the critical points to create test intervals: \((-\infty, -\frac{5}{4})\), \((-\frac{5}{4}, 0)\), \((0, 3)\), and \((3, \infty)\). Test a point from each interval in the factored inequality \( x(4x + 5)(x - 3) \leq 0 \).
05

Determine the sign of each interval

1. For \( x \in (-\infty, -\frac{5}{4}) \), let \( x = -2 \): \((-2)(4(-2) + 5)(-2 - 3) = (-2)(-3)(-5) = -30 \rightarrow negative\), 2. For \( x \in (-\frac{5}{4}, 0) \), let \( x = -1 \): \((-1)(4(-1) + 5)(-1 - 3) = (-1)(1)(-4) = 4 \rightarrow positive\), 3. For \( x \in (0, 3) \), let \( x = 1 \): \((1)(4(1) + 5)(1 - 3) = (1)(9)(-2) = -18 \rightarrow negative\), 4. For \( x \in (3, \infty) \), let \( x = 4 \): \((4)(4(4) + 5)(4 - 3) = (4)(21)(1) = 84 \rightarrow positive\)
06

Combine results

The inequality \( x(4x + 5)(x - 3) \leq 0 \) is satisfied where the polynomial is negative or zero. The sign analysis shows that the inequality holds in the intervals \((-\infty, -\frac{5}{4})\) and \((0, 3)\). Including the critical points where the polynomial is zero, the solution is: \[\boxed{(-\infty, -\frac{5}{4}] \cup [0, 3]}\]

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Key Concepts

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

Factoring Polynomials
Factoring polynomials is an essential step in solving polynomial inequalities. It involves expressing the polynomial as a product of simpler polynomials. In our exercise, the polynomial to factorize is \[4x^3 - 7x^2 - 15x \].
First, we factor out the common term, which is \(x\): \[ x(4x^2 - 7x - 15) \leq 0 \].
Next, we factor the quadratic expression inside the parentheses. We look for two numbers that multiply to \(4 \times (-15) = -60 \) and add up to \(-7\).
These numbers are \( -12 \text{ and } 5 \). Therefore, we can write: \( x(4x + 5)(x - 3) \leq 0 \).
Now the polynomial is completely factored!
Critical Points
Critical points are the values of \(x\) that make each factor in the polynomial zero. They are crucial in solving polynomial inequalities because they divide the number line into intervals that we can test for the inequality.
To find the critical points of \( x(4x + 5)(x - 3) \leq 0 \), we set each factor equal to zero and solve for \(x\):
  • \( x = 0 \)
  • \( 4x + 5 = 0 \rightarrow x = -\frac{5}{4} \)
  • \( x - 3 = 0 \rightarrow x = 3 \)
These points are \( x = 0 \), \( x = -\frac{5}{4} \), and \( x = 3 \).
They help us determine the intervals to test for the inequality.
Test Intervals
Test intervals help to determine where the polynomial expression \( x(4x + 5)(x - 3) \) is less than or equal to zero. We use the critical points to divide the number line into intervals:
  • \( (-\infty, -\frac{5}{4}) \)
  • \( (-\frac{5}{4}, 0) \)
  • \( (0, 3) \)
  • \( (3, \infty) \)
For each interval, we pick a test point and plug it into \( x(4x + 5)(x - 3) \) to determine if the result is positive or negative:
  • In \( (-\infty, -\frac{5}{4})\): Let \( x = -2 \). Test: \((-2)(4(-2) + 5)(-2 - 3) = -30 \rightarrow \text{ negative} \)
  • In \( (-\frac{5}{4}, 0) \): Let \( x = -1 \). Test: \((-1)(4(-1) + 5)(-1 - 3) = 4 \rightarrow \text{ positive} \)
  • In \( (0, 3) \): Let \( x = 1 \). Test: \((1)(4(1) + 5)(1 - 3) = -18 \rightarrow \text{ negative} \)
  • In \( (3, \infty) \): Let \( x = 4 \). Test: \((4)(4(4) + 5)(4 - 3) = 84 \rightarrow \text{ positive} \)
The inequality holds where the polynomial is negative or zero. Therefore, it is satisfied in the intervals: \((-\infty, -\frac{5}{4})\) and \((0, 3)\). Including the critical points, we get: \[ \boxed{(-\infty, -\frac{5}{4}] \text{ and } [0, 3]} \].

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