Question

Theorem: For any real number x,x ²−4∣x∣+16≥0 In a proof by cases of the theorem, there are three cases. The first case is that x≥4. The second case is that x≤−4. What is the third case? a. −4

Answer 1

The third case in the proof by cases of the theorem is when -4 < x < 4. This is because the given inequality is x² - 4|x| + 16 ≥ 0.

Explanation:

The third case in the proof by cases of the theorem is when -4 < x < 4. This is because the given inequality is x² - 4|x| + 16 ≥ 0. When x is between -4 and 4, the absolute value of x is positive, so we can simplify the inequality to x² - 4x + 16 ≥ 0. This inequality is true for all values of x, since the discriminant of the quadratic equation is negative, which means that the quadratic has no real roots. Therefore, the third case is -4 < x < 4.