International Baccalaureate Mathematics

Syllabus Content

Solutions of g(x) ≥ f(x), both graphically and analytically.

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Functions A relationship between two sets can be called a mapping. Elements of the first set (domain) are mapped to elements of the second set (range). A function is a special type of mapping for which one value in the domain maps to one, and only one value in the range.Pupils in Primary school will use the concept of function machines to perform calculations. They will then learn to ‘work backwards’ to find the inverse function. The study of functions becomes more formal as pupils become more proficient and able to cope with more complex mathematical ideas.

Furthermore

Official Guidance, clarification and syllabus links:

Graphical or algebraic methods for simple polynomials up to degree 3. Use of technology for these and other functions.

To find the solutions of the inequality $g(x) < f(x)$, you need to identify the values of $x$ that make $g(x)$ less than $f(x)$. Here are the general steps to follow:

Subtract $f(x)$ from both sides of the inequality to get $g(x) - f(x) < 0$.
Find the critical values of $x$ where $g(x) - f(x) = 0$. These values are where the inequality may change.
Test each interval between the critical values by choosing a test point within it and plugging it into the inequality. If the test point satisfies the inequality, then all points in that interval will satisfy the inequality. If it does not, then none of the points in that interval will satisfy the inequality.

Here are a couple of examples to illustrate this process:

Example 1:

Find the solutions of the inequality $x^2 - 3x < 2x - 1$.

Solution:

Subtracting $2x - 1$ from both sides gives $x^2 - 5x + 1 < 0$. Next, we find the critical values of $x$ by setting $x^2 - 5x + 1 = 0$ and solving for $x$ using the quadratic formula:

$$x = \frac{5 \pm \sqrt{21}}{2}$$

These values divide the real number line into three intervals: $(-\infty, \frac{5-\sqrt{21}}{2})$, $(\frac{5-\sqrt{21}}{2}, \frac{5+\sqrt{21}}{2})$, and $ (\frac{5+\sqrt{21}}{2}, \infty)$. Now we choose a test point in each interval and plug it into the inequality:

$-2$ does not satisfy $x^2 - 5x + 1 < 0$ for $(-\infty, \frac{5-\sqrt{21}}{2})$.
3 satisfies $x^2 - 5x + 1 < 0$ for $(\frac{5-\sqrt{21}}{2}, \frac{5+\sqrt{21}}{2})$.
6 does not satisfy $x^2 - 5x + 1 < 0$ for $ (\frac{5+\sqrt{21}}{2}, \infty)$.

Therefore, the solutions of the inequality are $ \frac{5-\sqrt{21}}{2} < x < \frac{5+\sqrt{21}}{2}$.

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