Question

question 15
solve the following rational inequality \\(\frac{x - 2}{x^2 - 36} > 0\\).
state your answer using interval notation. use u for union and oo for \\(\infty\\).
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Explanation:

Step1: Factor the denominator

First, factor the denominator \(x^2 - 36\) using the difference of squares formula \(a^2 - b^2=(a - b)(a + b)\). So, \(x^2 - 36=(x - 6)(x + 6)\). The rational inequality becomes \(\frac{x - 2}{(x - 6)(x + 6)}>0\).

Step2: Find critical points

The critical points are the values of \(x\) that make the numerator or the denominator equal to zero.

• For the numerator \(x - 2 = 0\), we get \(x = 2\).
• For the denominator \((x - 6)(x + 6)=0\), we get \(x = 6\) and \(x=- 6\).

These critical points divide the number line into the intervals: \((-\infty,-6)\), \((-6,2)\), \((2,6)\), and \((6,\infty)\).

Step3: Test intervals

We test a value from each interval in the inequality \(\frac{x - 2}{(x - 6)(x + 6)}>0\):

• For the interval \((-\infty,-6)\), let's choose \(x=-7\). Substitute into the inequality: \(\frac{-7 - 2}{(-7 - 6)(-7 + 6)}=\frac{-9}{(-13)(-1)}=\frac{-9}{13}<0\), so this interval does not satisfy the inequality.
• For the interval \((-6,2)\), let's choose \(x = 0\). Substitute: \(\frac{0 - 2}{(0 - 6)(0 + 6)}=\frac{-2}{(-6)(6)}=\frac{-2}{-36}=\frac{1}{18}>0\), so this interval satisfies the inequality.
• For the interval \((2,6)\), let's choose \(x = 3\). Substitute: \(\frac{3 - 2}{(3 - 6)(3 + 6)}=\frac{1}{(-3)(9)}=\frac{1}{-27}<0\), so this interval does not satisfy the inequality.
• For the interval \((6,\infty)\), let's choose \(x = 7\). Substitute: \(\frac{7 - 2}{(7 - 6)(7 + 6)}=\frac{5}{(1)(13)}=\frac{5}{13}>0\), so this interval satisfies the inequality.

We also need to note that \(x
eq - 6\) and \(x
eq6\) because they make the denominator zero (the function is undefined at these points) and \(x
eq2\) would make the numerator zero (but we are looking for where the expression is greater than zero, not equal to zero).

Answer:

\((-6,2)\cup(6,\infty)\)