Question

1. (3 points) find k so that the following function is continuous on any interval

f(x)=\begin{cases}kx^{2}-7&\text{for }x < 3\kx + 5&\text{for }xgeq3end{cases}
fully justify your answer.

Explanation:

Step1: Recall continuity condition

For a function to be continuous at $x = a$, $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)$. Here $a = 3$. So we need $\lim_{x
ightarrow 3^{-}}(kx^{2}-7)=\lim_{x
ightarrow 3^{+}}(kx + 5)$.

Step2: Calculate left - hand limit

$\lim_{x
ightarrow 3^{-}}(kx^{2}-7)=k(3)^{2}-7=9k - 7$.

Step3: Calculate right - hand limit

$\lim_{x
ightarrow 3^{+}}(kx + 5)=k(3)+5=3k + 5$.

Step4: Set left - hand and right - hand limits equal

Set $9k-7 = 3k + 5$.
Subtract $3k$ from both sides: $9k-3k-7=3k-3k + 5$, which simplifies to $6k-7 = 5$.
Add 7 to both sides: $6k-7 + 7=5 + 7$, getting $6k=12$.
Divide both sides by 6: $k=\frac{12}{6}=2$.

Answer:

$k = 2$