## Calculus with Applications (10th Edition)

$9-x^2\leq0$ $(3-x)(3+x)\leq0$ $-(x-3)(x+3)\leq0$ $(x-3)(x+3)\geq0$ $x \leq-3$ and $3\leq x$
Step 1: Factorize inequality (difference of squares) Step 2: Factorize $-1$ out of first Bracket (common factor) Step 3: Multiply with $-1$ on both sides of the inequality Step 4: Since the inequality was multiplied with a negative the direction of the inequality flips. Step 5: The inequality function is greater than $0$ for $x \leq-3$ and $3\leq x$