3. determine the point(s), if any, at which the function is discontinuous. classify any discontinuity as jump, removable, infinite, or other. (i) (\frac{x + 1}{x^2 + 1}) (ii) (\frac{x^2 - 9}{x^2 - 5x + 6}) (iii) (\frac{|x - 3|}{x - 3}) (iv) ( an(2x))

Question

3. determine the point(s), if any, at which the function is discontinuous. classify any discontinuity as jump, removable, infinite, or other. (i) (\frac{x + 1}{x^2 + 1}) (ii) (\frac{x^2 - 9}{x^2 - 5x + 6}) (iii) (\frac{|x - 3|}{x - 3}) (iv) (	an(2x))

Accepted Answer

### (i) For the function $ f(x) = \frac{x + 1}{x^2 + 1} $
# Explanation:
## Step 1: Analyze the denominator
The denominator is $ x^2 + 1 $. For real numbers $ x $, $ x^2\geq0 $, so $ x^2 + 1\geq1 $. The denominator is never zero for any real $ x $.
## Step 2: Conclusion about continuity
Since the denominator is non - zero for all real $ x $, the function $ f(x)=\frac{x + 1}{x^2+1} $ is continuous for all real numbers $ x\in\mathbb{R} $. There are no points of discontinuity.

### (ii) For the function $ f(x)=\frac{x^2 - 9}{x^2-5x + 6} $
# Explanation:
## Step 1: Factor numerator and denominator
Factor the numerator: $ x^2-9=(x - 3)(x + 3) $ (using the difference of squares formula $ a^2-b^2=(a - b)(a + b) $)
Factor the denominator: $ x^2-5x + 6=(x - 2)(x - 3) $ (by finding two numbers that multiply to 6 and add up to - 5, which are - 2 and - 3)
So the function can be written as $ f(x)=\frac{(x - 3)(x + 3)}{(x - 2)(x - 3)} $, $ x
eq3 $
## Step 2: Simplify the function (excluding $ x = 3 $)
Cancel out the common factor $ (x - 3) $ (for $ x
eq3 $), we get $ f(x)=\frac{x + 3}{x - 2} $, $ x
eq3 $
## Step 3: Analyze the denominator of the simplified function and the original function
- For the original function, the denominator is zero when $ x = 2 $ and $ x=3 $.
- For $ x = 2 $: The simplified function $ \frac{x + 3}{x - 2} $ has a denominator zero at $ x = 2 $, and $ \lim_{x
ightarrow2^+}f(x)=\lim_{x
ightarrow2^+}\frac{x + 3}{x - 2}=\infty $ and $ \lim_{x
ightarrow2^-}f(x)=\lim_{x
ightarrow2^-}\frac{x + 3}{x - 2}=-\infty $, so $ x = 2 $ is a point of infinite discontinuity.
- For $ x = 3 $: The limit $ \lim_{x
ightarrow3}f(x)=\lim_{x
ightarrow3}\frac{x + 3}{x - 2}=\frac{3 + 3}{3 - 2}=6 $. But the function is not defined at $ x = 3 $ in the original function. So $ x = 3 $ is a removable discontinuity.

### (iii) For the function $ f(x)=\frac{|x - 3|}{x - 3} $
# Explanation:
## Step 1: Analyze the absolute value function
Recall that $ |x-3|=\begin{cases}x - 3, & x\geq3\-(x - 3), & x<3\end{cases} $
## Step 2: Find the left - hand limit and right - hand limit at $ x = 3 $
- Right - hand limit ($ x
ightarrow3^+ $): When $ x
ightarrow3^+ $, $ |x - 3|=x - 3 $, so $ \lim_{x
ightarrow3^+}\frac{|x - 3|}{x - 3}=\lim_{x
ightarrow3^+}\frac{x - 3}{x - 3}=1 $
- Left - hand limit ($ x
ightarrow3^- $): When $ x
ightarrow3^- $, $ |x - 3|=-(x - 3) $, so $ \lim_{x
ightarrow3^-}\frac{|x - 3|}{x - 3}=\lim_{x
ightarrow3^-}\frac{-(x - 3)}{x - 3}=-1 $
## Step 3: Conclusion about discontinuity
Since $ \lim_{x
ightarrow3^+}f(x)
eq\lim_{x
ightarrow3^-}f(x) $, the function has a jump discontinuity at $ x = 3 $. Also, the function is not defined at $ x = 3 $.

### (iv) For the function $ y = \tan(2x) $
# Explanation:
## Step 1: Recall the domain of tangent function
We know that $ \tan\theta=\frac{\sin\theta}{\cos\theta} $, and $ \tan\theta $ is undefined when $ \cos\theta = 0 $. For $ y=\tan(2x) $, we set $ 2x=\frac{\pi}{2}+n\pi $, where $ n\in\mathbb{Z} $ (since $ \cos\theta = 0 $ when $ \theta=\frac{\pi}{2}+n\pi,n\in\mathbb{Z} $)
## Step 2: Solve for $ x $
Solve $ 2x=\frac{\pi}{2}+n\pi $ for $ x $, we get $ x=\frac{\pi}{4}+\frac{n\pi}{2},n\in\mathbb{Z} $
## Step 3: Analyze the discontinuity
At these points $ x = \frac{\pi}{4}+\frac{n\pi}{2},n\in\mathbb{Z} $, $ \cos(2x)=0 $, so $ \tan(2x) $ has vertical asymptotes. The limit $ \lim_{x
ightarrow\frac{\pi}{4}+\frac{n\pi}{2}^+}\tan(2x)=\infty $ and $ \lim_{x
ightarrow\frac{\pi}{4}+\frac{n\pi}{2}^-}\tan(2x)=-\infty $ (or vice - versa), so these are points of infinite discontinuity.

### (i)
# Answer:
The function $ \frac{x + 1}{x^2+1} $ is continuous for all real $ x $, no points of discontinuity.

### (ii)
# Answer:
- $ x = 2 $: Infinite discontinuity.
- $ x = 3 $: Removable discontinuity.

### (iii)
# Answer:
$ x = 3 $: Jump discontinuity.

### (iv)
# Answer:
$ x=\frac{\pi}{4}+\frac{n\pi}{2},n\in\mathbb{Z} $: Infinite discontinuity.

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QUESTION IMAGE

Question

Explanation:

(i) For the function \( f(x) = \frac{x + 1}{x^2 + 1} \)

Step 1: Analyze the denominator

Step 2: Conclusion about continuity

(ii) For the function \( f(x)=\frac{x^2 - 9}{x^2-5x + 6} \)

Step 1: Factor numerator and denominator

Step 2: Simplify the function (excluding \( x = 3 \))

Step 3: Analyze the denominator of the simplified function and the original function

(iii) For the function \( f(x)=\frac{|x - 3|}{x - 3} \)

Step 1: Analyze the absolute value function

Step 2: Find the left - hand limit and right - hand limit at \( x = 3 \)

Step 3: Conclusion about discontinuity

(iv) For the function \( y = \tan(2x) \)

Answer:

(ii)