Solve x²-36 < 0: Finding Values Where Function is Negative

Given the function:

$y=x^2-36$

Determine for which X values the following is true:

$f\left(x\right) < 0$

We want to find where the function $y = x^2 - 36$ is negative, expressed as $x^2 - 36 < 0$. To solve this, we start with solving the equation $x^2 - 36 = 0$ to find where the function is zero and change intervals from positive to negative or vice versa.

Let's solve $x^2 - 36 = 0$:

• $x^2 = 36$
• Taking the square root of both sides gives $x = \pm 6$.

So, the roots of the function are $x = -6$ and $x = 6$. The sign of the quadratic function changes at these points. To determine the intervals where $x^2 - 36 < 0$, we need to test the intervals determined by these roots: $(-\infty, -6)$, $(-6, 6)$, and $(6, \infty)$.

Next, we'll test these intervals:

• Choose a test point in $(-\infty, -6)$, for instance, $x = -7$:
• When $x = -7$, $x^2 - 36 = 49 - 36 = 13$ (positive).
• Choose a test point in $(-6, 6)$, for instance, $x = 0$:
• When $x = 0$, $x^2 - 36 = 0 - 36 = -36$ (negative).
• Choose a test point in $(6, \infty)$, for instance, $x = 7$:
• When $x = 7$, $x^2 - 36 = 49 - 36 = 13$ (positive).

From this analysis, we see that $x^2 - 36 < 0$ only in the interval $(-6, 6)$.

Therefore the values of $x$ for which $f(x) < 0$ are in the interval $-6 < x < 6$. This corresponds to choice 3 in the given options.

The solution to the problem is $-6 < x < 6$.