Answer to Question #301043 in Calculus for riyah

Question

Give me two examples with solutions of each types of discontinuities

Answer 1

Solution:

Removable discontinuity:

Example 1:

$f(x)=\frac{x^{2}-5 x+4}{x-4}$

Find the factors of the numerator and the denominator. The denominator cannot be factored further, but we can factor the numerator.

$\begin{gathered} =x^{2}-5 x+4 \\ =x^{2}-4 x-x+4 \\ =x(x-4)-1(x-4) \\ =(x-1)(x-4) \\ f(x)=\frac{(x-1)(x-4)}{x-4} \end{gathered}$

Now, we have identified the common factor which is x-4 . We need to set it equal to zero to get the removable discontinuity.

$\begin{aligned} &x-4=0 \\ &x=4 \end{aligned}$

Hence, the removable discontinuity of the function is at the point x=4 .

Example 2:

$f(x)=\frac{x^{2}-7 x+12}{x-4}$

Find the factors of the numerator and the denominator. The denominator cannot be factored further, but we can factor the numerator.

$\begin{gathered} =x^{2}-7 x+12 \\ =x^{2}-3 x-4x+12 \\ =x(x-3)-4(x-3) \\ =(x-3)(x-4) \\ f(x)=\frac{(x-3)(x-4)}{x-4} \end{gathered}$

Now, we have identified the common factor which is x-4 . We need to set it equal to zero to get the removable discontinuity.

$\begin{aligned} &x-4=0 \\ &x=4 \end{aligned}$

Hence, the removable discontinuity of the function is at the point x=4 .

Non-removable discontinuity:

Example 1:

Find the points of discontinuity of the $f(x)=\frac{1}{2 \sin x-1}$ .

Solution: Given that $f(x)=\frac{1}{2 \sin x-1}$ .

 f(x) is discontinuous when

$2 \sin x-1=0$

$\operatorname{Sin} x=1 / 2 \\x=\sin ^{-1}(1 / 2) \\x=\pi / 6 \text { (or) } 5 \pi / 6$

The general solution is $2 n \pi \pm \frac{\pi}{6} (or) 2 n \pi \pm \frac{5 \pi}{6}, \mathrm{n} \in \mathrm{z} .$

Example 2:

Find the points of discontinuity of the $f(x)=\frac{1}{2x+1}$ .

Solution: Given that $f(x)=\frac{1}{2x+1}$ .

 f(x) is discontinuous when

$2 x+1=0$

$2x=-1 \\x=\dfrac{-1}2$

f(x) is discontinuous at $x=\dfrac{-1}2$