Let f(x) = begin{cases} frac{x^3 + x^2 - 16x | vedclass

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(A) For $f(x)$ to be continuous at $x = 2$,we must have $\lim_{x 	o 2} f(x) = f(2) = k$. First,we evaluate the limit: $\lim_{x 	o 2} \frac{x^3 + x^2

Vedclass · Accepted Answer

$7$

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(A) For $f(x)$ to be continuous at $x = 2$,we must have $\lim_{x 	o 2} f(x) = f(2) = k$. First,we evaluate the limit: $\lim_{x 	o 2} \frac{x^3 + x^2 - 16x + 20}{(x - 2)^2}$. Since substituting $x = 2$ gives the indeterminate form $\frac{0}{0}$,we factor the numerator. By synthetic division or polynomial division,$x^3 + x^2 - 16x + 20 = (x - 2)^2(x + 5)$. Thus,$\lim_{x 	o 2} \frac{(x - 2)^2(x + 5)}{(x - 2)^2} = \lim_{x 	o 2} (x + 5) = 2 + 5 = 7$. Therefore,$k = 7$.

Let $f(x) = \begin{cases} \frac{x^3 + x^2 - 16x + 20}{(x - 2)^2}, & \text{if } x \neq 2 \\ k, & \text{if } x = 2 \end{cases}$. If $f(x)$ is continuous for all $x$,then $k =$

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