19 international practice exam ab mcq, part b - calculator required let f be the function defined above…

Explanation:

Step1: Recall continuity condition

A function $f(x)$ is continuous at $x = a$ if $\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a^{+}}f(x)=f(a)$. Here $a = 3$, so $\lim_{x\rightarrow3^{-}}(k^{3}+x)=\lim_{x\rightarrow3^{+}}\frac{16}{k^{2}-x}$.

Step2: Calculate left - hand limit

$\lim_{x\rightarrow3^{-}}(k^{3}+x)=k^{3}+3$.

Step3: Calculate right - hand limit

$\lim_{x\rightarrow3^{+}}\frac{16}{k^{2}-x}=\frac{16}{k^{2}-3}$.

Step4: Set up the equation

Set $k^{3}+3=\frac{16}{k^{2}-3}$. Cross - multiply to get $(k^{3}+3)(k^{2}-3)=16$. Expand the left - hand side: $k^{5}-3k^{3}+3k^{2}-9 = 16$, or $k^{5}-3k^{3}+3k^{2}-25=0$. We can try to solve this equation numerically. By using a graphing utility or a calculator with equation - solving capabilities, we find that there is no real positive value of $k$ that satisfies this equation.

Answer:

D. There is no such value of $k$.