Question

score on last try: 0 of 3 pts. see details for more. > next question get a similar question you can retry this question below 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 - 56}{x - 4} show that f(x) has a removable discontinuity at x = 4 and determine what value for f(4) would make f(x) continuous at x = 4. must define f(4)=21 x question help: video written example

Explanation:

Step1: Factor the numerator

Factor $2x^{2}+6x - 56$. First, factor out the common factor 2: $2(x^{2}+3x - 28)$. Then factor the quadratic inside the parentheses: $2(x + 7)(x - 4)$. So $f(x)=\frac{2(x + 7)(x - 4)}{x - 4}$.

Step2: Simplify the function

For $x
eq4$, we can cancel out the $(x - 4)$ terms. So $f(x)=2(x + 7)$ for $x
eq4$.

Step3: Find the limit as $x$ approaches 4

We find $\lim_{x
ightarrow4}f(x)=\lim_{x
ightarrow4}2(x + 7)$. Substitute $x = 4$ into $2(x + 7)$: $2(4+7)=2\times11 = 22$.

Step4: Determine the value to make the function continuous

To make $f(x)$ continuous at $x = 4$, we define $f(4)$ to be equal to $\lim_{x
ightarrow4}f(x)$. So we must define $f(4)=22$.

Answer:

$22$