Questions: find the x-value(s) (if any) at which f is not continuous. F(x)=(x-4)/(x^2-4x-32)

Transcript text: find the $x$-value(s) (if any) at which $f$ is not continuous. $F(x)=\frac{x-4}{x^{2}-4 x-32}$

Solution

Solution Steps

Step 1: Identify Points of Discontinuity

To find the points where the function $ F(x) = \frac{x-4}{x^2 - 4x - 32} $ is not continuous, we need to identify where the denominator is equal to zero, as division by zero is undefined.

Step 2: Solve for Zero in the Denominator

Set the denominator equal to zero and solve for $ x $:

\[ x^2 - 4x - 32 = 0 \]

This is a quadratic equation, which can be solved using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where $ a = 1 $, $ b = -4 $, and $ c = -32 $.

\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-32)}}{2 \cdot 1} \]

\[ x = \frac{4 \pm \sqrt{16 + 128}}{2} \]

\[ x = \frac{4 \pm \sqrt{144}}{2} \]

\[ x = \frac{4 \pm 12}{2} \]

This gives us two solutions:

\[ x = \frac{4 + 12}{2} = 8 \]

\[ x = \frac{4 - 12}{2} = -4 \]

Step 3: Conclusion on Continuity

The function $ F(x) $ is not continuous at $ x = 8 $ and $ x = -4 $ because these values make the denominator zero, leading to undefined points in the function.