Solving Inequalities of the Form g(x) ≥ f(x)
In Mathematics AA HL, students are expected to solve inequalities of the form g(x) ≥ f(x), where g(x) and f(x) are functions. This type of inequality compares two functions and asks for the values of x where one function is greater than or equal to the other.
Graphical Approach
One method to solve g(x) ≥ f(x) is by graphing both functions and identifying the regions where g(x) is above or on f(x).
Consider the inequality $x^2 + 1 \geq 2x$. To solve this graphically:
- Graph $y = x^2 + 1$ (a parabola)
- Graph $y = 2x$ (a straight line)
- Identify where the parabola is above or touching the line
The solution is the x-values where the parabola is above or touching the line, which in this case is $x \leq -1$ or $x \geq 2$.
When graphing, pay attention to the points of intersection between g(x) and f(x). These points often represent the boundaries of the solution set.
Analytical Approach
The analytical method involves algebraically manipulating the inequality to solve for an exact value of $x$.
- Rearrange the inequality to have zero on one side: g(x) - f(x) ≥ 0
- Solve this new inequality using algebraic techniques, such as factoring
For $x^2 + 1 \geq 2x$:
- Rearrange: $x^2 - 2x + 1 \geq 0$
- This is a quadratic inequality. We can solve it by finding the roots of $x^2 - 2x + 1 = 0$
- The roots are x = 1 ± 1
- The parabola opens upward, so the solution is $x \leq -1$ or $x \geq 2$
The analytical method often requires factoring, completing the square, or using the quadratic formula for quadratic inequalities.
Algebraic Methods for Simple Polynomials
For polynomials up to degree 3, students should be familiar with various algebraic techniques:
Linear Inequalities
These are the simplest form, solved by isolating x on one side.
Solve $2x + 3 > 7$
- Subtract 3 from both sides: $2x > 4$
- Divide both sides by 2: $x > 2$
Quadratic Inequalities
These involve squaring and can be solved by factoring or using the quadratic formula.
When multiplying by a negative number or taking the reciprocla, it is necessary to flip the inequality sign.
While it is true that $2<6$, it is also true than $-2 > -6$, and $\frac{1}{2} > \frac{1}{6}$.
