3. determine the point(s), if any, at which the function is discontinuous. classify any discontinuity as…
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) (\tan(2x))
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^{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} )
Explanation:
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\neq3,2 )
Step 2: Simplify the function
Cancel out the common factor ( (x - 3) ) (for ( x\neq3 )), we get ( f(x)=\frac{x + 3}{x - 2} ), ( x\neq3,2 )
Step 3: Analyze discontinuities
- At ( x = 3 ): The original function is undefined at ( x = 3 ), but the limit ( \lim_{x\rightarrow3}\frac{(x - 3)(x + 3)}{(x - 2)(x - 3)}=\lim_{x\rightarrow3}\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\rightarrow2^{-}}\frac{x + 3}{x - 2}=-\infty ) and ( \lim_{x\rightarrow2^{+}}\frac{x + 3}{x - 2}=+\infty ). So, ( x = 2 ) is an infinite discontinuity.
(iii) For the function ( f(x)=\frac{|x - 3|}{x - 3} )
Explanation:
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\rightarrow3^{+}}f(x)=1 ) and ( \lim_{x\rightarrow3^{-}}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) )
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=(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\rightarrow(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.