To find and classify the discontinuities of the function [tex]\( h(x) = \frac{x^2 - 64}{x - 8} \)[/tex], we will follow these steps:
### Step 1: Factor the numerator
The first step is to factor the numerator [tex]\( x^2 - 64 \)[/tex]. Notice that [tex]\( x^2 - 64 \)[/tex] is a difference of squares and can be factored as follows:
[tex]\[
x^2 - 64 = (x - 8)(x + 8)
\][/tex]
### Step 2: Simplify the function
Now we can rewrite the function [tex]\( h(x) \)[/tex] using the factored form of the numerator:
[tex]\[
h(x) = \frac{(x - 8)(x + 8)}{x - 8}
\][/tex]
Next, we can cancel the common factor [tex]\( x - 8 \)[/tex] in the numerator and the denominator, but we must note that this cancellation is valid only when [tex]\( x \neq 8 \)[/tex]:
[tex]\[
h(x) = x + 8 \quad \text{for} \quad x \neq 8
\][/tex]
### Step 3: Identify discontinuities
We see that the original function [tex]\( h(x) = \frac{x^2 - 64}{x - 8} \)[/tex] is undefined at [tex]\( x = 8 \)[/tex] because the denominator becomes zero.
#### Removable discontinuity
To determine if the discontinuity at [tex]\( x = 8 \)[/tex] is removable, we examine the simplified form [tex]\( h(x) = x + 8 \)[/tex]. If the function can be redefined at [tex]\( x = 8 \)[/tex] to make it continuous, then the discontinuity is removable.
Consider the simplified function:
[tex]\[
h(x) = x + 8
\][/tex]
Let's evaluate the simplified function at [tex]\( x = 8 \)[/tex]:
[tex]\[
h(8) = 8 + 8 = 16
\][/tex]
Thus, we can redefine the original function at [tex]\( x = 8 \)[/tex] to be 16, making the function continuous. Therefore, the discontinuity at [tex]\( x = 8 \)[/tex] is removable.
#### Nonremovable discontinuities
From the simplified function, we can see that there are no other values of [tex]\( x \)[/tex] that cause the denominator to be zero, aside from the already addressed [tex]\( x = 8 \)[/tex]. Thus, there are no nonremovable discontinuities for this function.
### Conclusion
The function [tex]\( h(x) = \frac{x^2 - 64}{x - 8} \)[/tex] has a removable discontinuity at [tex]\( x = 8 \)[/tex] and no nonremovable discontinuities.
Therefore, the answers are:
[tex]\[
\text{Removable: } c = 8
\][/tex]
[tex]\[
\text{Nonremovable: } c = \text{None}
\][/tex]
