Question

1. for what values of a and b is

$g(x)=\

$$\begin{cases}-2, & x\\leq -1 \\\\ ax - b, & -1 < x < 1 \\\\ 3, & x\\geq 1\\end{cases}$$

$
continuous at every x?

Explanation:

Step1: Use the definition of continuity at \(x = - 1\)

For a function \(y = g(x)\) to be continuous at \(x=-1\), \(\lim_{x
ightarrow - 1^{-}}g(x)=\lim_{x
ightarrow - 1^{+}}g(x)=g(-1)\)

We know that \(\lim_{x
ightarrow - 1^{-}}g(x)=-2\), \(g(-1)=-2\) and \(\lim_{x
ightarrow - 1^{+}}g(x)=a\times(-1)-b\)

So, \(-a - b=-2\), which can be rewritten as \(a + b=2\)

Step2: Use the definition of continuity at \(x = 1\)

For a function \(y = g(x)\) to be continuous at \(x = 1\), \(\lim_{x
ightarrow1^{-}}g(x)=\lim_{x
ightarrow1^{+}}g(x)=g(1)\)

We know that \(\lim_{x
ightarrow1^{+}}g(x)=3\), \(g(1) = 3\) and \(\lim_{x
ightarrow1^{-}}g(x)=a\times1-b\)

So, \(a - b=3\)

Step3: Solve the system of equations

We have the system of equations \(

$$\begin{cases}a + b=2\\a - b=3\end{cases}$$

\)

Add the two equations: \((a + b)+(a - b)=2 + 3\)

\(2a=5\), then \(a=\frac{5}{2}\)

Substitute \(a=\frac{5}{2}\) into \(a + b=2\): \(\frac{5}{2}+b=2\)

\(b=2-\frac{5}{2}=-\frac{1}{2}\)

Answer:

\(a=\frac{5}{2}\) and \(b =-\frac{1}{2}\)