Question

1. what value(s) of k would make the function continuous?

g(x)=\begin{cases}-6x^{2}+18x, & xleq1\\k^{2}-k, & x > 1end{cases}

1. given the function: g(x)=\begin{cases}2x - 5, & xleq - 1\\x^{2}+2, & x > - 1end{cases}

a.) find (g(-2))
b.) find (g(0))

Explanation:

Step1: Recall the condition for continuity

For a function to be continuous at $x = a$, $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)$. For $g(x)$ at $x = 1$, we need $\lim_{x
ightarrow 1^{-}}g(x)=\lim_{x
ightarrow 1^{+}}g(x)$.

Step2: Calculate the left - hand limit

When $x
ightarrow 1^{-}$, $g(x)=-6x^{2}+18x$. Substitute $x = 1$ into $-6x^{2}+18x$: $-6(1)^{2}+18(1)=-6 + 18=12$.

Step3: Calculate the right - hand limit

When $x
ightarrow 1^{+}$, $g(x)=k^{2}-k$. Set $k^{2}-k = 12$. Rearrange to get a quadratic equation $k^{2}-k - 12=0$.

Step4: Solve the quadratic equation

Factor the quadratic equation $k^{2}-k - 12=(k - 4)(k+3)=0$. So $k=4$ or $k=-3$.

Step5: Solve for $g(-2)$ in part A of question 6

Since $-2\leq - 1$, for $g(x)$ when $x=-2$, use $g(x)=2x - 5$. Substitute $x=-2$: $g(-2)=2(-2)-5=-4 - 5=-9$.

Step6: Solve for $g(0)$ in part B of question 6

Since $0>-1$, for $g(x)$ when $x = 0$, use $g(x)=x^{2}+2$. Substitute $x = 0$: $g(0)=0^{2}+2=2$.

Answer:

For question 4: $k = 4$ or $k=-3$
For question 6 A: $g(-2)=-9$
For question 6 B: $g(0)=2$