Question

function $f(x)$ is said to have a removable discontinuity at $x = a$ if: $f$ is either not defined or not continuous at $x = a$. $f(a)$ could either be defined or re - defined so that the new function is continuous at $x=a$. let $f(x)=\frac{2x^{2}+4x - 70}{x - 5}$. show that $f(x)$ has a removable discontinuity at $x = 5$ and determine what value for $f(5)$ would make $f(x)$ continuous at $x = 5$. must define $f(5)=$. question help: message instructor

Explanation:

Step1: Factor the numerator

Factor $2x^{2}+4x - 70$ as $2(x^{2}+2x - 35)=2(x + 7)(x - 5)$. So $f(x)=\frac{2(x + 7)(x - 5)}{x - 5}$, $x
eq5$.

Step2: Simplify the function

Cancel out the common factor $(x - 5)$ (for $x
eq5$), we get $f(x)=2(x + 7)$ for $x
eq5$.

Step3: Find the limit as $x$ approaches 5

$\lim_{x
ightarrow5}f(x)=\lim_{x
ightarrow5}2(x + 7)$. Substitute $x = 5$ into $2(x + 7)$, we have $2(5+7)=24$.

Answer:

$24$