Solve (x-6)(x+6) > 0: Finding Values Where Function is Positive

Look at the following function:

$y=\left(x-6\right)\left(x+6\right)$

Determine for which values of $x$ the following is true:

$f(x) > 0$

To solve this problem, we need to find the values of $x$ that make $(x-6)(x+6) > 0$.

Let's consider the critical points where each factor could change sign by setting each factor equal to zero:

• $x-6 = 0$ gives $x = 6$.
• $x+6 = 0$ gives $x = -6$.

These critical points divide the number line into three intervals:

• $x < -6$
• $-6 < x < 6$
• $x > 6$

We will test each interval to see where $f(x) = (x-6)(x+6) > 0$:

1. Interval $x < -6$:

If $x < -6$, both $(x-6)$ and $(x+6)$ are negative (e.g., test $x = -7$).
$(x-6) \cdot (x+6) = (-)\cdot(-) > 0$: The product is positive.

2. Interval $-6 < x < 6$:

If $-6 < x < 6$, $(x-6)$ is negative, and $(x+6)$ is positive (e.g., test $x = 0$).
$(x-6) \cdot (x+6) = (-)\cdot(+) < 0$: The product is negative.

3. Interval $x > 6$:

If $x > 6$, both $(x-6)$ and $(x+6)$ are positive (e.g., test $x = 7$).
$(x-6) \cdot (x+6) = (+)\cdot(+) > 0$: The product is positive.

Therefore, the inequality $(x-6)(x+6) > 0$ holds for $x < -6$ or $x > 6$. Thus, the correct answer is:

$x > 6$ or $x < -6$