Answer :
To determine the intervals on which the given function [tex]\( f(x) = \frac{x-3}{x^2-9} \)[/tex] is continuous, we need to find where the function is undefined, as these points are where the function might have discontinuities.
### Step-by-Step Solution:
1. Simplify the Denominator:
The denominator of the function [tex]\( f(x) \)[/tex] is [tex]\( x^2 - 9 \)[/tex]. Let's factorize it:
[tex]\[ x^2 - 9 = (x - 3)(x + 3) \][/tex]
2. Identify Points of Discontinuity:
The function [tex]\( f(x) \)[/tex] will be undefined wherever the denominator is zero. Thus, we solve for where [tex]\( (x - 3)(x + 3) = 0 \)[/tex]:
[tex]\[ x - 3 = 0 \quad \text{or} \quad x + 3 = 0 \][/tex]
[tex]\[ x = 3 \quad \text{or} \quad x = -3 \][/tex]
Therefore, the function is undefined at [tex]\( x = 3 \)[/tex] and [tex]\( x = -3 \)[/tex].
3. Intervals of Continuity:
The function will be continuous on all intervals of the real line except at the points of discontinuity identified above, which are [tex]\( x = 3 \)[/tex] and [tex]\( x = -3 \)[/tex].
As such, the function [tex]\( f(x) \)[/tex] will be continuous on the following intervals:
[tex]\[ (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \][/tex]
### Summary:
The function [tex]\( f(x) = \frac{x-3}{x^2-9} \)[/tex] is continuous on the intervals:
[tex]\[ (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \][/tex]
It is discontinuous at [tex]\( x = 3 \)[/tex] and [tex]\( x = -3 \)[/tex].
### Step-by-Step Solution:
1. Simplify the Denominator:
The denominator of the function [tex]\( f(x) \)[/tex] is [tex]\( x^2 - 9 \)[/tex]. Let's factorize it:
[tex]\[ x^2 - 9 = (x - 3)(x + 3) \][/tex]
2. Identify Points of Discontinuity:
The function [tex]\( f(x) \)[/tex] will be undefined wherever the denominator is zero. Thus, we solve for where [tex]\( (x - 3)(x + 3) = 0 \)[/tex]:
[tex]\[ x - 3 = 0 \quad \text{or} \quad x + 3 = 0 \][/tex]
[tex]\[ x = 3 \quad \text{or} \quad x = -3 \][/tex]
Therefore, the function is undefined at [tex]\( x = 3 \)[/tex] and [tex]\( x = -3 \)[/tex].
3. Intervals of Continuity:
The function will be continuous on all intervals of the real line except at the points of discontinuity identified above, which are [tex]\( x = 3 \)[/tex] and [tex]\( x = -3 \)[/tex].
As such, the function [tex]\( f(x) \)[/tex] will be continuous on the following intervals:
[tex]\[ (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \][/tex]
### Summary:
The function [tex]\( f(x) = \frac{x-3}{x^2-9} \)[/tex] is continuous on the intervals:
[tex]\[ (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \][/tex]
It is discontinuous at [tex]\( x = 3 \)[/tex] and [tex]\( x = -3 \)[/tex].
