Question

question 9
a function $f(x)$ is said to have a removable discontinuity at $x=a$ if:

1. $f$ is either not defined or not continuous at $x=a$.
2. $f(a)$ could either be defined or redefined so that the new function is continuous at $x=a$.

let $f(x)=\frac{2x^{2}+6x-8}{x-1}$
show that $f(x)$ has a removable discontinuity at $x=1$ and determine what value for $f(1)$ would make $f(x)$ continuous at $x=1$.
must define $f(1)=$

Explanation:

Step1: Check continuity at $x=1$

At $x=1$, the denominator $x-1=0$, so $f(1)$ is undefined. Now check the limit as $x\to1$.

Step2: Factor the numerator

Factor $2x^2+6x-8$:
$2x^2+6x-8=2(x^2+3x-4)=2(x+4)(x-1)$

Step3: Simplify the function

Cancel $(x-1)$ for $x
eq1$:
$f(x)=\frac{2(x+4)(x-1)}{x-1}=2(x+4)$ for $x
eq1$

Step4: Compute the limit as $x\to1$

$\lim_{x\to1}f(x)=\lim_{x\to1}2(x+4)=2(1+4)=10$
Since the limit exists but $f(1)$ is undefined, the discontinuity is removable. Defining $f(1)$ equal to this limit makes the function continuous.

Answer:

$10$