Question

find k so that the following function is continuous.
f(x)=\begin{cases}\frac{5x^{4}-20x^{3}}{x - 4}&x
eq4\\k&x = 4end{cases}
enter dne if no such k - value is possible.
k=

Explanation:

Step1: Simplify the rational - function for $x

eq4$
Factor the numerator of $\frac{5x^{4}-20x^{3}}{x - 4}$. We have $5x^{4}-20x^{3}=5x^{3}(x - 4)$. So, $\frac{5x^{4}-20x^{3}}{x - 4}=\frac{5x^{3}(x - 4)}{x - 4}=5x^{3}$ for $x
eq4$.

Step2: Find the limit as $x$ approaches 4

We find $\lim_{x
ightarrow4}\frac{5x^{4}-20x^{3}}{x - 4}=\lim_{x
ightarrow4}5x^{3}$. Substitute $x = 4$ into $5x^{3}$. Then $5x^{3}\big|_{x = 4}=5\times4^{3}=5\times64 = 320$.

Step3: Use the definition of continuity

For the function $f(x)$ to be continuous at $x = 4$, we must have $\lim_{x
ightarrow4}f(x)=f(4)$. Since $\lim_{x
ightarrow4}\frac{5x^{4}-20x^{3}}{x - 4}=320$ and $f(4)=k$, then $k = 320$.

Answer:

$320$