Question

question 19 (5 points)
find (a) and (b) so that
g(x)=\begin{cases}ax^{2},&xleq2\ax + b,&2<xleq4\bx - 1,&x>4end{cases}
is continuous.

Explanation:

Step1: Use continuity at $x = 2$

For the function to be continuous at $x = 2$, the left - hand limit and the right - hand limit must be equal.
The left - hand limit as $x\to2^{-}$ is $\lim_{x\to2^{-}}g(x)=\lim_{x\to2^{-}}ax^{2}=4a$.
The right - hand limit as $x\to2^{+}$ is $\lim_{x\to2^{+}}g(x)=\lim_{x\to2^{+}}(ax + b)=2a + b$.
So, $4a=2a + b$, which simplifies to $2a - b=0$.

Step2: Use continuity at $x = 4$

For the function to be continuous at $x = 4$, the left - hand limit and the right - hand limit must be equal.
The left - hand limit as $x\to4^{-}$ is $\lim_{x\to4^{-}}g(x)=\lim_{x\to4^{-}}(ax + b)=4a + b$.
The right - hand limit as $x\to4^{+}$ is $\lim_{x\to4^{+}}g(x)=\lim_{x\to4^{+}}(bx-1)=4b - 1$.
So, $4a + b=4b - 1$, which simplifies to $4a-3b=-1$.

Step3: Solve the system of equations

We have the system of equations:

$$\begin{cases}2a - b = 0\\4a-3b=-1\end{cases}$$

From the first equation $b = 2a$.
Substitute $b = 2a$ into the second equation:
$4a-3(2a)=-1$.
$4a - 6a=-1$.
$-2a=-1$, so $a=\frac{1}{2}$.
Since $b = 2a$, then $b = 1$.

Answer:

$a=\frac{1}{2}$, $b = 1$