Question

1. for what value of k is $h(x)=\begin{cases}kx^{2}-6&x < 2\\4x + 10&xgeq2end{cases}$ continuous at $x = 2$?

Explanation:

Step1: Recall continuity condition

For a function to be continuous at $x = a$, $\lim_{x
ightarrow a^{-}}h(x)=\lim_{x
ightarrow a^{+}}h(x)=h(a)$. Here $a = 2$, so $\lim_{x
ightarrow 2^{-}}(kx^{2}-6)=\lim_{x
ightarrow 2^{+}}(4x + 10)$.

Step2: Calculate left - hand limit

$\lim_{x
ightarrow 2^{-}}(kx^{2}-6)=k(2)^{2}-6=4k - 6$.

Step3: Calculate right - hand limit

$\lim_{x
ightarrow 2^{+}}(4x + 10)=4\times2+10=8 + 10=18$.

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

Set $4k-6 = 18$. Add 6 to both sides: $4k=18 + 6=24$. Then divide both sides by 4: $k=\frac{24}{4}=6$.

Answer:

$k = 6$