Question

For what range of values of \(x\), will the inequality \(15x - \frac{2}{x} > 1\) hold?

• A. \(x > 0.4\)
• B. \(x < \frac{1}{3}\)
• C. \(-\frac{1}{3} < x < 0.4, \, x > \frac{15}{2}\)
• D. \(-\frac{1}{3} < x < 0, \, x > \frac{2}{5}\)

Hint:

Move everything to one side, combine over a common denominator, factor the quadratic, then run a sign chart across the three critical points -1/3, 0 and 2/5.

Answer 1

The Correct Option is D

A second way to confirm this is to test sample values from each candidate range directly in the original inequality $15x - \frac{2}{x} > 1$, rather than working with the factored fraction.

1. Test $x = -0.2$, which lies in $-\frac{1}{3} < x < 0$: $15(-0.2) - \frac{2}{-0.2} = -3 + 10 = 7$. Since $7 > 1$, this point satisfies the inequality.
2. Test $x = 1$, which lies in $x > \frac{2}{5}$: $15(1) - \frac{2}{1} = 15 - 2 = 13$. Since $13 > 1$, this point also satisfies the inequality.
3. Test $x = 0.2$, which lies in $0 < x < \frac{2}{5}$: $15(0.2) - \frac{2}{0.2} = 3 - 10 = -7$. Since $-7 > 1$ is false, this region does not satisfy the inequality, so it must be excluded from the answer.
4. Test $x = -1$, which lies in $x < -\frac{1}{3}$: $15(-1) - \frac{2}{-1} = -15 + 2 = -13$. Since $-13 > 1$ is false, this region is also excluded.
5. Test $x = 5$, checking further into $x > \frac{2}{5}$ to make sure the region does not close off again: $15(5) - \frac{2}{5} = 75 - 0.4 = 74.6 > 1$, true, so the entire region $x > \frac{2}{5}$ stays valid with no upper cutoff.

These sample points confirm the solution set is $-\frac{1}{3} < x < 0$ combined with $x > \frac{2}{5}$, matching the sign-chart result and ruling out option (3), which wrongly treats $0 < x < 0.4$ as part of the solution.

\[\boxed{-\frac{1}{3} < x < 0,\ x > \frac{2}{5}}\]