Answer
$\left[-\frac{1}{3}, \frac{1}{2}\right] \cup [1, \infty)$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a polynomial function.
$x^2(7-6x) \leq 1$,
$7x^2-6x^3-1 \leq0$,
$-6x^3+7x^2-1 \leq 0$,
$(x-1)(x-\frac{1}{2})(-6x-2) \leq 0$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$(x-1)(x-\frac{1}{2})(-6x-2)=0$
$x=1$ or $x=\frac{1}{2}$ or $x=-\frac{1}{3}$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \leq 0 ? \\
& &(a-1)(a-\frac{1}{2})(-6a-2)& \\
(-\infty, -\frac{1}{3}) & -2 & (-)(-)(+) & F\\
(-\frac{1}{3}, \frac{1}{2}) & 0 & (-)(-)(-) & T\\
(\frac{1}{2}, 1) & 0.6 & (-)(+)(-) &F\\
(1,\infty) & 2 & (+)(+)(-) & T
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $\left[-\frac{1}{3}, \frac{1}{2}\right] \cup [1, \infty)$
