12. $$ f(x)=\left\{\begin{array}{ll}\frac{x^{2}-9}{x+3}, & x
eq-3 \ 3, & x=-3\end{array} ight. $$ at $$ \mathrm{c}=-3 $$ In Problems $$ 13-20 $$, find all points of discontinuity. $$ \begin{array}{ll} ext { 13. } g(x)=\frac{\left(x^{2}-1 ight)^{2}}{5} & ext { 14. } f(x)=0 \ ext { 15. } f(x)=\frac{x^{2}+6 x+9}{x^{2}+2 x-15} & ext { 16. } g(x)=\frac{x-3}{x^{2}+x} \ \begin{array}{ll} ext { 17. } h(x)=\frac{x-7}{x^{3}-x} & ext { 18. } f(x)=\frac{x}{x} \ ext { 19. } p(x)=\frac{x}{x^{2}+1} & ext { 20. } f(x)=\frac{x^{4}}{x^{4}-1}\end{array}\end{array} $$.

Question

12. $$ f(x)=\left\{\begin{array}{ll}\frac{x^{2}-9}{x+3}, & x 
eq-3 \ 3, & x=-3\end{array}ight. $$ at $$ \mathrm{c}=-3 $$ In Problems $$ 13-20 $$, find all points of discontinuity. $$ \begin{array}{ll}	ext { 13. } g(x)=\frac{\left(x^{2}-1ight)^{2}}{5} & 	ext { 14. } f(x)=0 \ 	ext { 15. } f(x)=\frac{x^{2}+6 x+9}{x^{2}+2 x-15} & 	ext { 16. } g(x)=\frac{x-3}{x^{2}+x} \ \begin{array}{ll}	ext { 17. } h(x)=\frac{x-7}{x^{3}-x} & 	ext { 18. } f(x)=\frac{x}{x} \ 	ext { 19. } p(x)=\frac{x}{x^{2}+1} & 	ext { 20. } f(x)=\frac{x^{4}}{x^{4}-1}\end{array}\end{array} $$.

UpStudy AI · Accepted Answer

To find the points of discontinuity for each function, we will analyze them one by one. A function is discontinuous at a point if it is not defined at that point or if the limit does not equal the function value at that point.

### Problem 12: 
**Function:** 
$$
f(x)=\left\{\begin{array}{ll}\frac{x^{2}-9}{x+3}, & x \neq-3 \ 3, & x=-3\end{array}\right.
$$

**Analysis:**
1. The function is defined for all $$ x $$ except $$ x = -3 $$ in the first case.
2. We can simplify $$ \frac{x^{2}-9}{x+3} $$ for $$ x \neq -3 $$:
   $$
   \frac{x^{2}-9}{x+3} = \frac{(x-3)(x+3)}{x+3} = x-3 \quad (x \neq -3)
   $$
3. The limit as $$ x $$ approaches -3:
   $$
   \lim_{x \to -3} f(x) = -3 - 3 = -6
   $$
4. Since $$ f(-3) = 3 $$ and the limit does not equal the function value, there is a discontinuity at $$ x = -3 $$.

### Problem 13: 
**Function:** 
$$
g(x)=\frac{\left(x^{2}-1\right)^{2}}{5}
$$

**Analysis:**
1. This function is a polynomial divided by a constant, which is defined for all $$ x $$.
2. There are no points of discontinuity.

### Problem 14: 
**Function:** 
$$
f(x)=0
$$

**Analysis:**
1. This is a constant function, which is defined for all $$ x $$.
2. There are no points of discontinuity.

### Problem 15: 
**Function:** 
$$
f(x)=\frac{x^{2}+6x+9}{x^{2}+2x-15}
$$

**Analysis:**
1. Factor the denominator:
   $$
   x^{2}+2x-15 = (x+5)(x-3)
   $$
2. The function is undefined at $$ x = -5 $$ and $$ x = 3 $$.
3. Therefore, there are discontinuities at $$ x = -5 $$ and $$ x = 3 $$.

### Problem 16: 
**Function:** 
$$
g(x)=\frac{x-3}{x^{2}+x}
$$

**Analysis:**
1. Factor the denominator:
   $$
   x^{2}+x = x(x+1)
   $$
2. The function is undefined at $$ x = 0 $$ and $$ x = -1 $$.
3. Therefore, there are discontinuities at $$ x = 0 $$ and $$ x = -1 $$.

### Problem 17: 
**Function:** 
$$
h(x)=\frac{x-7}{x^{3}-x}
$$

**Analysis:**
1. Factor the denominator:
   $$
   x^{3}-x = x(x^{2}-1) = x(x-1)(x+1)
   $$
2. The function is undefined at $$ x = 0, 1, -1 $$.
3. Therefore, there are discontinuities at $$ x = 0, 1, -1 $$.

### Problem 18: 
**Function:** 
$$
f(x)=\frac{x}{x}
$$

**Analysis:**
1. This function is defined for all $$ x $$ except $$ x = 0 $$.
2. Therefore, there is a discontinuity at $$ x = 0 $$.

### Problem 19: 
**Function:** 
$$
p(x)=\frac{x}{x^{2}+1}
$$

**Analysis:**
1. The denominator $$ x^{2}+1 $$ is never zero for any real $$ x $$.
2. Therefore, there are no points of discontinuity.

### Problem 20: 
**Function:** 
$$
f(x)=\frac{x^{4}}{x^{4}-1}
$$

**Analysis:**
1. Factor the denominator:
   $$
   x^{4}-1 = (x^{2}-1)(x^{2}+1) = (x-1)(x+1)(x^{2}+1)
   $$
2. The function is undefined at $$ x = 1 $$ and $$ x = -1 $$.
3. Therefore, there are discontinuities at $$ x = 1 $$ and $$ x = -1 $$.

### Summary of Points of Discontinuity:
- **12:** $$ x = -3 $$
- **13:** None
- **14:** None
- **15:** $$ x = -5, 3 $$
- **16:** $$ x = 0, -1 $$
- **17:** $$ x = 0, 1, -1 $$
- **18:** $$ x = 0 $$
- **19:** None
- **20:** $$ x = 1, -1 $$
Points of discontinuity: - Problem 12: $$ x = -3 $$ - Problem 13: None - Problem 14: None - Problem 15: $$ x = -5, 3 $$ - Problem 16: $$ x = 0, -1 $$ - Problem 17: $$ x = 0, 1, -1 $$ - Problem 18: $$ x = 0 $$ - Problem 19: None - Problem 20: $$ x = 1, -1 $$

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