📘 Module 1.3: Intervals and Inequalities

Definition of order in ℝ, intervals, and linear and quadratic inequalities, with examples, theorems, and diagrams.

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1 Order in \(\mathbb{R}\)

📖 Definition — Strict Order

Let \(a,b \in \mathbb{R}\): 1. \(a < b \iff b-a > 0\); 2. \(a > b \iff a-b > 0\).

1.1 Examples

• \(3<7\), since \(7-3=4>0\).

• \(-5<-2\), since \(-2-(-5)=3>0\).

Generic diagram:

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2 Non-Strict Inequalities

📖 Definition — Non-Strict Order

Let \(a,b \in \mathbb{R}\):

1. \(a \leqslant b \iff (a<b) \text{ or } (a=b)\);
2. \(a \geqslant b \iff (a>b) \text{ or } (a=b)\).

• \(a<b\) and \(a>b\): strict inequalities.
  
• \(a\leqslant b\) and \(a\geqslant b\): non-strict inequalities.

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3 Fundamental Theorems of Order

📐 Theorem 1 — Signs

1. \(a>0 \iff a\) is positive.
   
2. \(a<0 \iff a\) is negative.

📐 Theorem 2 — Transitivity

If \(a<b\) and \(b<c\), then \(a<c\).

📐 Theorem 3 — Addition and Multiplication

1. If \(a<b\), then \(a+c<b+c\).
   
2. If \(a<b\) and \(c>0\), then \(ac<bc\).
   
3. If \(a<b\) and \(c<0\), then \(ac>bc\).

📐 Theorem 4 — Sum of Inequalities

If \(a<b\) and \(c<d\), then \(a+c<b+d\).

4 Chains of Inequalities

A number \(x\) is between \(a\) and \(b\) if \(a<x<b\).
Example: \(2<x<5\).

Other forms:
- \(a \leqslant x \leqslant b\) → closed interval;
- \(a \leqslant x < b\), \(a < x \leqslant b\) → half-open intervals.

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5 Intervals

5.1 Open interval

\[(a,b) = \{x\in\mathbb{R}\mid a<x<b\}\]

5.2 Closed interval

\[[a,b] = \{x\in\mathbb{R}\mid a\leqslant x\leqslant b\}\]

5.3 Half-open intervals

• \((a,b] = \{x \mid a<x\leqslant b\}\)
  

• \([a,b) = \{x \mid a\leqslant x<b\}\)

5.4 Infinite intervals

• \((a,+\infty) = \{x \mid x>a\}\)
  

• \((-\infty,b) = \{x \mid x<b\}\)
  

• \([a,+\infty) = \{x \mid x\geqslant a\}\)

• \((-\infty,b] = \{x \mid x\leqslant b\}\)

• \((-\infty,+\infty)=\mathbb{R}\)

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6 Linear Inequalities

General form (with \(a \neq 0\)):

\[ \boxed{\, ax+b < c \,} \]

\[ \boxed{\, ax+b \leqslant c \,} \]

\[ \boxed{\, ax+b > c \,} \]

\[ \boxed{\, ax+b \geqslant c \,} \]

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📌 Important note:
- If \(a=0\), the inequality reduces to a constant statement (\(b<c\), \(b\leqslant c\), etc.), with no variable.
- If \(a\neq 0\), we can isolate \(x\):

• For \(a>0\), the order is preserved.
  
• For \(a<0\), the inequality reverses its direction.

🧮 Examples — Linear Inequalities

6.1 Cases with \(a>0\)

1. \(2x-3<5 \;\;\Rightarrow\;\; x<4\)
   Solution: \((-\infty,4)\).

2. \(3x+1\leqslant 7 \;\;\Rightarrow\;\; x\leqslant 2\)
   Solution: \((-\infty,2]\).

3. \(5x-4>11 \;\;\Rightarrow\;\; x>3\)
   Solution: \((3,+\infty)\).

4. \(4x+2\geqslant 10 \;\;\Rightarrow\;\; x\geqslant 2\)
   Solution: \([2,+\infty)\).

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6.2 Cases with \(a<0\)

5. \(-2x+3<7 \;\;\Rightarrow\;\; x>-2\)
   Solution: \((-2,+\infty)\).

6. \(-3x+1\leqslant -8 \;\;\Rightarrow\;\; x\geqslant 3\)
   Solution: \([3,+\infty)\).

7. \(-4x+5>1 \;\;\Rightarrow\;\; x<1\)
   Solution: \((-\infty,1)\).

8. \(-x-2\geqslant 5 \;\;\Rightarrow\;\; x\leqslant -7\)
   Solution: \((-\infty,-7]\).

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📌 Key summary:
- If \(a>0\), the inequality keeps its direction.
- If \(a<0\), the inequality reverses its direction.

🧮 Worked Examples in Detail

• Example 1: \(2x-3<5\) (coefficient \(a>0\))

\[ \begin{aligned} 2x-3 &< 5 && \text{(given)}\\ 2x &< 5+3 = 8 && \text{(add 3 to both sides)}\\ x &< \frac{8}{2} = 4 && \text{(divide by 2>0, inequality preserved)} \end{aligned} \]

✅ Solution (interval): \((-\infty,4)\).
📌 Check: \(x=3 \Rightarrow 3<5\) (true); \(x=4 \Rightarrow 5\not<5\) (false).

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• Example 5: \(-2x+3<7\) (coefficient \(a<0\))

\[ \begin{aligned} -2x+3 &< 7 && \text{(given)}\\ -2x &< 7-3 = 4 && \text{(subtract 3 from both sides)}\\ x &> \frac{4}{-2} = -2 && \text{(divide by -2<0, \textbf{reverse} the inequality)} \end{aligned} \]

✅ Solution (interval): \((-2,+\infty)\).
📌 Check: \(x=0 \Rightarrow 3<7\) (true); \(x=-2 \Rightarrow 7\not<7\) (false).

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Teaching summary:
- Adding/subtracting: inequality preserved.
- Multiplying/dividing by a positive number: inequality preserved.
- Multiplying/dividing by a negative number: inequality reversed.

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7 Quadratic Inequalities

General form (with \(a \neq 0\)):

\[ \boxed{ax^2 +bx + c < 0} \]

\[ \boxed{ax^2 +bx + c > 0} \]

\[ \boxed{ax^2 +bx + c \leqslant 0} \]

\[ \boxed{ax^2 +bx + c \geqslant 0} \]

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Solve by analyzing the parabola \(y=ax^2+bx+c\).

📌 Example 1: \(x^2-5x+6>0\)
- Roots: \(x=2,3\)
- Parabola opens upwards.
- Positive sign outside of \((2,3)\).
- Solution: \((-\infty,2)\cup(3,+\infty)\).

Diagram on the real line:

Parabola graph:

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📌 Example 1: \(x^2-5x+6>0\)

🧮 Detailed Solution

1. Identification: quadratic inequality with \(a=1>0\), \(b=-5\), \(c=6\).

2. Associated equation:
   \[ x^2 - 5x + 6 = 0 \]

3. Discriminant (Δ):
   \[ \Delta = b^2 - 4ac = (-5)^2 - 4\cdot 1 \cdot 6 = 25 - 24 = 1 \]

4. Real roots:
   \[ x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{5 \pm 1}{2} \] Thus, \(x_1=2\) and \(x_2=3\).

5. Sketch of the parabola: since \(a=1>0\), the parabola is concave upward.

6. Sign of the parabola:

   • For \(x<2\), the parabola is above the axis (\(f(x)>0\)).
     
   • For \(2<x<3\), the parabola is below the axis (\(f(x)<0\)).
     
   • For \(x>3\), the parabola is again above the axis (\(f(x)>0\)).

7. Conclusion (solution of the inequality \(f(x)>0\)):
   \[ x \in (-\infty,2)\cup(3,+\infty) \]

Diagram on the real line:

Parabola graph and sign analysis:

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📌 Example 2: \(-x^2+4x-3 \geqslant 0\)

🧮 Detailed Solution

1. Identification: quadratic inequality with \(a=-1<0\), \(b=4\), \(c=-3\).

2. Associated equation:
   \[ -x^2+4x-3=0 \]

3. Discriminant (Δ):
   \[ \Delta = b^2-4ac = (4)^2 - 4\cdot(-1)\cdot(-3) = 16 - 12 = 4 \]

4. Real roots:
   \[ x=\frac{-b\pm\sqrt{\Delta}}{2a} = \frac{-4\pm 2}{-2} \] Thus, \(x_1=1\) and \(x_2=3\).

5. Sketch of the parabola: since \(a=-1<0\), the parabola is concave downward.
   This means the parabola is above the \(x\)-axis between the roots and below it outside.

6. Sign of the parabola:

   • For \(x<1\), \(f(x)<0\).
     
   • For \(1\leqslant x\leqslant 3\), \(f(x)\geqslant 0\).
     
   • For \(x>3\), \(f(x)<0\) again.

7. Conclusion (solution of the inequality \(f(x)\geqslant 0\)):
   \[ x \in [1,3] \]

Diagram on the real line:

Parabola graph and sign analysis:

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8 Product Inequalities

General form:

\[ \boxed{(x-a)(x-b) < 0} \]

\[ \boxed{(x-a)(x-b) > 0} \]

\[ \boxed{(x-a)(x-b) \leqslant 0} \]

\[ \boxed{(x-a)(x-b) \geqslant 0} \]

The solution is determined by the sign of the factors in each interval defined by the roots.

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📌 Example 1: Positive product
Solve: \[ x(x-1) > 0 \]

Detailed solution

• Roots: \(x=0\) and \(x=1\).
  
• Intervals: \((-\infty,0),\ (0,1),\ (1,+\infty)\).
  
• Sign test:
  • For \(x<0\): both factors negative ⇒ product positive.
    
  • For \(0<x<1\): one positive, one negative ⇒ product negative.
    
  • For \(x>1\): both positive ⇒ product positive.

✅ Solution: \((-\infty,0)\cup(1,+\infty)\).

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📌 Example 2: Negative product
Solve: \[ (x-1)(x+2) < 0 \]

Detailed solution

• Roots: \(x=-2\) and \(x=1\).
  
• Intervals: \((-\infty,-2),\ (-2,1),\ (1,+\infty)\).
  
• Sign test:
  • For \(x<-2\): both negative ⇒ product positive.
    
  • For \(-2<x<1\): different signs ⇒ product negative.
    
  • For \(x>1\): both positive ⇒ product positive.

✅ Solution: \((-2,1)\).

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9 Quotient Inequalities

General form:

\[ \boxed{\dfrac{f(x)}{g(x)} < 0,\quad g(x)\neq 0} \]

\[ \boxed{\dfrac{f(x)}{g(x)} > 0,\quad g(x)\neq 0} \]

\[ \boxed{\dfrac{f(x)}{g(x)} \leqslant 0,\quad g(x)\neq 0} \]

\[ \boxed{\dfrac{f(x)}{g(x)} \geqslant 0,\quad g(x)\neq 0} \]

The solution is determined by the sign chart of numerator and denominator.

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9.1 📌 What are Asymptotes?

📖 Definitions

Let a rational function \(f(x)=\dfrac{p(x)}{q(x)}\).

• Vertical asymptote:
  We say that \(x=a\) is a vertical asymptote if:
  • \(q(a)=0\) (the denominator vanishes at \(a\)); and
  • \(p(a)\ne0\) (the numerator does not vanish at \(a\)).
    In this case, \(f(x)\) is not defined at \(x=a\) and the graph of the function approaches a vertical line at \(x=a\).
• Horizontal asymptote:
  We say that \(y=L\) is a horizontal asymptote if \[ \lim_{x\to\pm\infty} f(x) = L. \] This means that, for very large positive or negative values of \(x\), the graph of \(f(x)\) approaches the horizontal line \(y=L\).

📌 These concepts will be explored in detail in the Module on Rational Functions and, more rigorously, in the Module on Limits.

🔎 Remark

• In problems involving rational inequalities, the values that annihilate the denominator never belong to the solution, since the expression becomes undefined.
  
• Therefore, we treat these points as boundaries in the sign analysis, and call them vertical asymptotes.
  
• Horizontal asymptotes help describe the behavior at infinity of a rational function.

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📌 Example 1: Positive quotient
Solve: \[ \frac{x+1}{x-3} > 0 \]

Detailed solution

• Zeros: numerator \(x=-1\).
  
• Asymptote: denominator \(x=3\).
  
• Intervals: \((-\infty,-1),\ (-1,3),\ (3,+\infty)\).
  
• Signs:
  • \((-\infty,-1)\): both negative ⇒ quotient positive.
    
  • \((-1,3)\): numerator positive, denominator negative ⇒ quotient negative.
    
  • \((3,+\infty)\): both positive ⇒ quotient positive.

✅ Solution: \((-\infty,-1)\cup(3,+\infty)\).

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📌 Example 2: Quotient with more factors
Solve: \[ \frac{x^2-1}{x^2-3x} > 0 \]

Detailed solution

\[ \frac{(x-1)(x+1)}{x(x-3)} > 0, \quad x\neq 0,3. \]

• Zeros/asymptotes: \(-1,\,0,\,1,\,3\).
  
• Intervals: \((-\infty,-1),\ (-1,0),\ (0,1),\ (1,3),\ (3,\infty)\).
  
• Sign test:
  • \((-\infty,-1)\): positive.
    
  • \((-1,0)\): negative.
    
  • \((0,1)\): positive.
    
  • \((1,3)\): negative.
    
  • \((3,\infty)\): positive.

✅ Solution: \((-\infty,-1)\cup(0,1)\cup(3,\infty)\).

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📝 Proposed Exercises

1. Represent on the real line:
   1. \((1,5]\)
      
   2. \((-\infty,-2)\)
      
   3. \([-3,3]\)
2. Solve:
   1. \(3x+1\geqslant 7\)
      
   2. \(x^2-4\leqslant 0\)
      
   3. \((x-1)(x-4)<0\)

✅ Answer Key

1. 1. open dot at 1, closed dot at 5
      
   2. up to \(-2\) (open)
      
   3. closed from \(-3\) to \(3\)
2. 1. \([2,+\infty)\)
      
   2. \([-2,2]\)
      
   3. \((1,4)\)

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9.2 🧠 Review Exercises

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Exercise 1 — Compound inequality and interval

Solve and write the solution in interval notation: \[ 1<2x+3\le 9 \]

Step-by-step solution

\[ \begin{aligned} 1<2x+3\le 9 &\iff 1-3<2x\le 9-3 \\ &\iff -2<2x\le 6 \\ &\iff -1<x\le 3 \quad (\text{divide by }2>0) \end{aligned} \] Solution: \((-1,3]\).

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Exercise 2 — Negative coefficient (sign inversion)

Solve: \[ -3x+5\ge 2 \]

Step-by-step solution

\[ \begin{aligned} -3x+5\ge 2 &\iff -3x\ge -3 \\ &\iff x\le 1 \quad (\text{divide by }-3<0 \Rightarrow \text{flip inequality}) \end{aligned} \] Solution: \((-\infty,1]\).

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Exercise 3 — Factorable quadratic inequality (< 0)

Solve: \[ x^2-x-6<0 \]

Step-by-step solution

Factoring: \(x^2-x-6=(x-3)(x+2)\).
Roots: \(x=-2\) and \(x=3\). Since \(a=1>0\), the parabola is concave upwards.
Thus, \(f(x)<0\) between the roots: \[ x\in(-2,3). \] Solution: \((-2,3)\).

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Exercise 4 — Quadratic inequality (non-integer roots) (≥ 0)

Solve: \[ 2x^2+3x-2\ge 0 \]

Step-by-step solution

Discriminant: \(\Delta=b^2-4ac=3^2-4\cdot2\cdot(-2)=9+16=25\).
Roots: \[ x=\frac{-3\pm\sqrt{25}}{2\cdot2}=\frac{-3\pm5}{4}\;\Rightarrow\;x_1=-2,\;x_2=\tfrac12. \] With \(a=2>0\), the parabola is concave upwards; so \(f(x)\ge 0\) outside the interval between the roots (and includes the roots because it is \(\geqslant\)): \[ x\in(-\infty,-2]\cup\left[\tfrac12,\infty\right). \] Solution: \((-\infty,-2]\cup[0.5,\infty)\).

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Exercise 5 — Operations with intervals

Let \(A=(-\infty,4)\) and \(B=[1,7)\).
(a) \(A\cap B\) (b) \(A\cup B\)

Step-by-step solution

• \(A\cap B\): points that are in both.
  \(A\) goes up to \(4\) (open), \(B\) starts at \(1\) (closed) and goes up to \(7\) (open).
  Intersection: from \(1\) to \(4\), including \(1\) and excluding \(4\): \([1,4)\).

• \(A\cup B\): points that are in at least one.
  The union covers from \(-\infty\) up to \(7\) open (since \(B\) already covers \([1,7)\) and \(A\) covers \((-\infty,4)\)): \((-\infty,7)\).

Answers: (a) \([1,4)\) (b) \((-\infty,7)\).

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Exercise 6 — Solve the inequalities below.

1. Product: \(x(x-1)>0\)

Step-by-step solution

Roots at \(x=0\) and \(x=1\). Since the leading coefficient is \(+1\), the product is positive outside the interval between the roots.
✅ Solution: \((-\infty,0)\cup(1,\infty)\).

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2. Product: \((x-1)(x+2)<0\)

Step-by-step solution

Roots at \(x=-2\) and \(x=1\). Leading coefficient \(+1\) ⇒ negative between the roots.
✅ Solution: \((-2,1)\).

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3. Quadratic: \(x^2+4x-21>0\)

Step-by-step solution

\(\Delta=4^2-4\cdot1\cdot(-21)=100\).
Roots: \(x=\tfrac{-4\pm10}{2}\Rightarrow x=-7,\,3\).
Parabola opens upwards ⇒ \(>0\) outside.
✅ Solution: \((-\infty,-7)\cup(3,\infty)\).

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4. Quadratic: \(2x^2+x<3\)

Step-by-step solution

\(2x^2+x-3<0\).
\(\Delta=1+24=25\).
Roots: \(x=\tfrac{-1\pm5}{4}\Rightarrow x=-\tfrac32,\,1\).
\(<0\) between the roots.
✅ Solution: \(\left(-\tfrac32,\,1\right)\).

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5. Quadratic: \(4x^2+10x-6<0\)

Step-by-step solution

\(\Delta=10^2-4\cdot4\cdot(-6)=196\).
Roots: \(x=\tfrac{-10\pm14}{8}\Rightarrow x=-3,\,\tfrac12\).
\(<0\) between the roots.
✅ Solution: \((-3,\,\tfrac12)\).

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6. Quadratic: \(x^2+2x+4>0\)

Step-by-step solution

\(\Delta=2^2-4\cdot1\cdot4=-12<0\) and \(a=1>0\) ⇒ the expression is always positive.
✅ Solution: \(\mathbb{R}\).

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Exercise 7 — Real radicals (radicand ≥ 0)

1. \(\sqrt{4-x^2}\)

Step-by-step solution

We require \(4-x^2\ge0 \Rightarrow -2\le x\le 2\).
✅ Solution: \([-2,2]\).

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2. \(\dfrac{1}{\sqrt{4-3x}}\)

Step-by-step solution

The radicand must be positive (and the denominator \(\ne0\)): \(4-3x>0 \Rightarrow x<\tfrac43\).
✅ Solution: \((-\infty,\tfrac43)\).

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3. \(\dfrac{1}{\sqrt{x^2-x-12}}\)

Step-by-step solution

We require \(x^2-x-12>0\Rightarrow(x-4)(x+3)>0\) ⇒ outside the roots.
✅ Solution: \((-\infty,-3)\cup(4,\infty)\).

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Exercise 8 — Positivity of rational functions

1. \(\dfrac{x}{x^2+4}>0\)

Step-by-step solution

Denominator \(x^2+4>0\) for all \(x\). Sign of the fraction = sign of \(x\).
✅ Solution: \((0,\infty)\).

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2. \(\dfrac{x}{x^2-4}>0\)

Step-by-step solution

Critical points: \(x=0,\pm2\) (undefined at \(\pm2\)). Sign chart ⇒ positive on \((-2,0)\) and \((2,\infty)\).
✅ Solution: \((-2,0)\cup(2,\infty)\).

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3. \(\dfrac{x+1}{x-3}>0\)

Step-by-step solution

Zero: \(-1\). Asymptote: \(3\). Sign by intervals ⇒ \((-\infty,-1)\) and \((3,\infty)\).
✅ Solution: \((-\infty,-1)\cup(3,\infty)\).

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4. \(\dfrac{x^2-1}{x^2-3x}>0\)

Step-by-step solution

\(\dfrac{(x-1)(x+1)}{x(x-3)}>0\), with \(x\ne0,3\). Critical points: \(-1,0,1,3\).
Signs ⇒ \((-\infty,-1)\cup(0,1)\cup(3,\infty)\).
✅ Solution: \((-\infty,-1)\cup(0,1)\cup(3,\infty)\).

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Exercise 9 — Describe and diagram the set (in \(\mathbb{R}\))

1. \(x^2<4\)

Step-by-step solution

\(-2<x<2\).
✅ Solution: \((-2,2)\).

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2. \(x^2+3x-4>0\)

Step-by-step solution

\((x+4)(x-1)>0\) ⇒ outside the roots.
✅ Solution: \((-\infty,-4)\cup(1,\infty)\).

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3. \(x^2+6x+8\le0\)

Step-by-step solution

\((x+2)(x+4)\le0\) ⇒ between the roots, including them.
✅ Solution: \([-4,-2]\).

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4. \(x^2<5x+14\)

Step-by-step solution

\(x^2-5x-14<0\Rightarrow(x-7)(x+2)<0\) ⇒ between the roots.
✅ Solution: \((-2,7)\).

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5. \(2x^2>x+6\)

Step-by-step solution

\(2x^2-x-6>0\). \(\Delta=49\). Roots: \(-\tfrac32\) and \(2\).
\(>0\) outside.
✅ Solution: \((-\infty,-\tfrac32)\cup(2,\infty)\).

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6. \(6x^2+13x<5\)

Step-by-step solution

\(6x^2+13x-5<0\). \(\Delta=289\). Roots: \(-\tfrac52\) and \(\tfrac13\).
\(<0\) between.
✅ Solution: \((-\tfrac52,\,\tfrac13)\).

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7. \(x^3+3x^2>10x\)

Step-by-step solution

\(x(x^2+3x-10)=x(x+5)(x-2)>0\).
Sign chart with critical points \(-5,0,2\) ⇒
✅ Solution: \((-5,0)\cup(2,\infty)\).

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Exercise 10 — Solve

1. \(-4<2-x<7\)

Step-by-step solution

Adding \(-2\): \(-6<-x<5\). Multiplying by \(-1\) (inequality flips): \(6>x>-5\).
✅ Solution: \((-5,6)\).

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2. \(\dfrac{2x-1}{x}<3\)

Step-by-step solution

\[ \frac{2x-1}{x}-3=\frac{-x-1}{x}<0 \ \Longleftrightarrow\ \frac{x+1}{x}>0. \] Critical points: \(-1,0\). Signs ⇒ positive on \((-\infty,-1)\) and \((0,\infty)\).
✅ Solution: \((-\infty,-1)\cup(0,\infty)\).

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3. \(\dfrac{x}{x+2}<1\)

Step-by-step solution

\[ \frac{x}{x+2}-1=\frac{-2}{x+2}<0 \ \Longleftrightarrow\ \frac{2}{x+2}>0 \] (numerator \(>0\)) ⇒ \(x+2>0 \Rightarrow x>-2\), with \(x\ne-2\).
✅ Solution: \((-2,\infty)\).

10 🔎 Summary Sheet — Module 1.3 (Order, Intervals, and Inequalities)

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📖 Order Definitions in \(\mathbb{R}\)

• \(a<b \iff b-a>0\); \(a>b \iff a-b>0\).
• \(a\le b \iff (a<b \text{ or } a=b)\); \(a\ge b \iff (a>b \text{ or } a=b)\).
• Strict inequalities: \(<,>\). Non-strict: \(\le,\ge\).

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📏 Order Properties — Quick use

• Addition: \(a<b \Rightarrow a+c<b+c\).
• Product with \(c>0\): inequality preserved.
• Product with \(c<0\): inequality reversed.
• Transitivity: if \(a<b\) and \(b<c\), then \(a<c\).

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📦 Intervals and Notation

• Open: \((a,b)=\{x\mid a<x<b\}\).
• Closed: \([a,b]=\{x\mid a\le x\le b\}\).
• Half-open: \((a,b],\ [a,b)\).
• Infinite: \((a,\infty),\ (-\infty,b),\ [a,\infty),\ (-\infty,b]\).

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🧮 Linear Inequalities (with (a))

General form: \(\boxed{\,ax+b\ \{<,\le,>,\ge\}\ c\,}\)

Step-by-step

1. Isolate the term \(ax\) (add/subtract on both sides).
   
2. Divide by \(a\). If \(a<0\), flip the inequality.
   
3. Write the solution in interval notation and test a value if useful.
   

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🟦 Quadratic Inequalities

• Study the sign of \(f(x)=ax^2+bx+c\) using roots and concavity.
• \(\Delta=b^2-4ac\).
  
• If \(a>0\): \(f(x)\) is \(>0\) outside the roots and \(<0\) between.
  
• If \(a<0\): behavior reversed.

Step-by-step

1. Solve \(ax^2+bx+c=0\) (factor/Quadratic Formula).
   
2. Mark the roots on the number line and note concavity (sign of \(a\)).
   
3. Select the intervals satisfying \(>,\ge,<,\le\).
   
4. Include the roots if the inequality is non-strict (\(\ge,\le\)).
   

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✖️ Product Inequalities

• Type: \((x-a)(x-b)\ \{<,>,\le,\ge\}\ 0\).
  
• Rule: build a sign chart of factors; the product is \(+\) if an even number of negatives, \(−\) if odd.

Quick example

\(x(x-1)>0 \Rightarrow (-\infty,0)\cup(1,\infty)\).
(Roots \(0,1\); signs: \(+,-,+\); we want \(+\) → outside the interval.)

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➗ Quotient Inequalities

• Type: \(\dfrac{f(x)}{g(x)}\ \{<,>,\le,\ge\}\ 0\), with \(g(x)\ne0\).
  
• Rule: signs are split by zeros of \(f\) (numerator) and asymptotes \(g=0\) (excluded!).
  
• Quotient is \(+\) if numerator and denominator share the same sign; \(−\) if signs are opposite.

Quick example

\(\dfrac{x+1}{x-3}>0 \Rightarrow (-\infty,-1)\cup(3,\infty)\).
(Criticals \(-1\) [zero] and \(3\) [asymptote]; signs: \(+,-,+\); we want \(+\).)

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🧰 Quick Checklist to Solve

• Identify type: linear / quadratic / product / quotient.
  
• Find critical points (roots, asymptotes).
  
• Build a sign chart by intervals.
  
• Apply the sign rule requested (\(>,\ge,<,\le\)).
  
• Decide inclusion/exclusion of boundaries.
  
• Write in interval notation and test a value.

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⚠️ Common Mistakes to Avoid

• Forgetting to flip the inequality when dividing by a negative number.
  
• Including asymptotes (\(g(x)=0\)) in quotients.
  
• Not checking if the root is included when the relation is \(\ge\) or \(\le\).
  
• Losing critical points when simplifying common factors (note restrictions first!).
  

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