Shane04 wrote:
Bunuel wrote:
Which of the following is a solution set to the equation: \(x(x-1)(x-6)<0\) ?
A. \(1<x ≤ 6\)
B. \(x ≤ 0\)
C. \(x < 0\)
D. \(6 ≤ x\)
E. \(0 ≤ x\)
Bunuel can you explain the solution to getting C?
\(x(x-1)(x-6)<0\)
Here is how to solve the above inequality easily. The "roots", in ascending order, are 0, 1, and 6, which gives us 4 ranges:
\(x < 0\);
\(0 < x < 1\);
\(1 < x < 6\);
\(x > 6\).
Next, test an extreme value for \(x\): if \(x\) is some large enough number, say 10, then all three multiples will be positive, giving a positive result for the whole expression. So when \(x > 6\), the expression is positive. Now the trick: as in the 4th range, the expression is positive, then in the 3rd it'll be negative, in the 2nd it'll be positive, and finally in the 1st, it'll be negative: \(\text{(- + - +)}\). So, the ranges when the expression is negative are: \(x < 0\) and \(1 < x < 6\).
P.S. You can apply this technique to any inequality of the form \((ax - b)(cx - d)(ex - f)... > 0\) or \( < 0\). If one of the factors is in the form \((b - ax)\) instead of \((ax - b)\), simply rewrite it as \(-(ax - b)\) and adjust the inequality sign accordingly. For example, consider the inequality \((4 - x)(2 + x) > 0\). Rewrite it as \(-(x - 4)(2 + x) > 0\). Multiply by -1 and flip the sign: \((x - 4)(2 + x) < 0\). The "roots", in ascending order, are -2 and 4, giving us three ranges: \(x < -2\), \(-2 < x < 4\), and \(x > 4\). If \(x\) is some large enough number, say 100, both factors will be positive, yielding a positive result for the entire expression. Thus, when \(x > 4\), the expression is positive. Now, apply the alternating sign trick: as the expression is positive in the 3rd range, it will be negative in the 2nd range, and positive in the 1st range: \(\text{(+ - +)}\). Therefore, the ranges where the expression is negative are: \(-2 < x < 4\).
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9. Inequalities
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