Question

1. 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) (\tan(2x))

Explanation:

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

Step 1: Analyze the denominator

The denominator is \( x^{2}+1 \). For real numbers, \( x^{2}\geq0 \), so \( x^{2}+1\geq1>0 \) for all real \( x \).

Step 2: Determine continuity

Since the denominator is never zero for any real \( x \), the function is defined for all real numbers. A rational function is continuous everywhere in its domain. So, this function has no points of discontinuity.

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

Step 1: Factor numerator and denominator

• Numerator: \( x^{2}-9=(x - 3)(x + 3) \) (using \( a^{2}-b^{2}=(a - b)(a + b) \))
• Denominator: \( x^{2}-5x + 6=(x - 2)(x - 3) \) (factoring quadratic)

So, \( f(x)=\frac{(x - 3)(x + 3)}{(x - 2)(x - 3)} \), \( x
eq3,2 \)

Step 2: Simplify the function

Cancel out the common factor \( (x - 3) \) (for \( x
eq3 \)), we get \( f(x)=\frac{x + 3}{x - 2} \), \( x
eq3,2 \)

Step 3: Analyze discontinuities

• At \( x = 3 \): The original function is undefined at \( x = 3 \), but the limit \( \lim_{x

ightarrow3}\frac{(x - 3)(x + 3)}{(x - 2)(x - 3)}=\lim_{x
ightarrow3}\frac{x + 3}{x - 2}=\frac{3 + 3}{3 - 2}=6 \) exists. So, \( x = 3 \) is a removable discontinuity.

• At \( x = 2 \): The denominator of the simplified function \( \frac{x + 3}{x - 2} \) is zero. \( \lim_{x

ightarrow2^{-}}\frac{x + 3}{x - 2}=-\infty \) and \( \lim_{x
ightarrow2^{+}}\frac{x + 3}{x - 2}=+\infty \). So, \( x = 2 \) is an infinite discontinuity.

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

Step 1: Analyze the absolute value function

We know that \( |x - 3|=

$$\begin{cases}x - 3, & x\geq3\\-(x - 3), & x<3\end{cases}$$

\)

Step 2: Find the domain and analyze the function

The function is undefined at \( x = 3 \) (denominator is zero).

• For \( x>3 \): \( f(x)=\frac{x - 3}{x - 3}=1 \)
• For \( x<3 \): \( f(x)=\frac{-(x - 3)}{x - 3}=-1 \)

Step 3: Analyze the limit at \( x = 3 \)

\( \lim_{x
ightarrow3^{+}}f(x)=1 \) and \( \lim_{x
ightarrow3^{-}}f(x)=-1 \). Since the left - hand limit and right - hand limit exist but are not equal, \( x = 3 \) is a jump discontinuity.

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

Answer:

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 \( 2x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z} \)

Step 2: Solve for \( x \)

\( x=(2n + 1)\frac{\pi}{4},n\in\mathbb{Z} \)

Step 3: Analyze the discontinuity

At \( x=(2n + 1)\frac{\pi}{4} \), \( \cos(2x)=0 \), so \( \tan(2x) \) is undefined. \( \lim_{x
ightarrow(2n + 1)\frac{\pi}{4}^{\pm}}\tan(2x)=\pm\infty \) (depending on the side of the limit). So, \( x=(2n + 1)\frac{\pi}{4},n\in\mathbb{Z} \) are infinite discontinuities.

(i) Answer: No points of discontinuity.

(ii) Answer: Removable discontinuity at \( x = 3 \), infinite discontinuity at \( x = 2 \)

(iii) Answer: Jump discontinuity at \( x = 3 \)

(iv) Answer: Infinite discontinuities at \( x=(2n + 1)\frac{\pi}{4},n\in\mathbb{Z} \)