Question

for what values of a and b is the following function continuous at every x?

$f(x)=\

$$\begin{cases} -8 & x\\leq -3 \\\\ ax - b & -3 < x < 3 \\\\ 13 & x\\geq 3 \\end{cases}$$

$

find the values of a and b for which the function $f$ continuous at every $x$.

$a = \square$ and $b = \square$
(type integers or simplified fractions.)

Explanation:

Step1: Check continuity at \( x = -3 \)

For a function to be continuous at \( x = -3 \), the left - hand limit, right - hand limit, and the function value at \( x=-3 \) must be equal.
The left - hand limit as \( x\to - 3^{-} \): Since for \( x\leq - 3\), \( f(x)=-8 \), so \( \lim_{x\to - 3^{-}}f(x)=-8 \).
The right - hand limit as \( x\to - 3^{+} \): For \( - 3\lt x\lt3\), \( f(x)=ax - b \), so \( \lim_{x\to - 3^{+}}f(x)=a(-3)-b=-3a - b \).
The function value at \( x = - 3 \) is \( f(-3)=-8 \).
So, we have the equation: \( -3a - b=-8 \).

Step2: Check continuity at \( x = 3 \)

For a function to be continuous at \( x = 3 \), the left - hand limit, right - hand limit, and the function value at \( x = 3 \) must be equal.
The left - hand limit as \( x\to 3^{-} \): For \( - 3\lt x\lt3\), \( f(x)=ax - b \), so \( \lim_{x\to 3^{-}}f(x)=a(3)-b = 3a - b \).
The right - hand limit as \( x\to 3^{+} \): Since for \( x\geq3\), \( f(x)=13 \), so \( \lim_{x\to 3^{+}}f(x)=13 \).
The function value at \( x = 3 \) is \( f(3)=13 \).
So, we have the equation: \( 3a - b=13 \).

Step3: Solve the system of equations

We have the system of equations:
\[

$$\begin{cases} -3a - b=-8\\ 3a - b=13 \end{cases}$$

\]
Subtract the first equation from the second equation:
\((3a - b)-(-3a - b)=13-(-8)\)
\(3a - b + 3a + b=13 + 8\)
\(6a=21\)
\(a=\frac{21}{6}=\frac{7}{2}\)

Substitute \( a = \frac{7}{2} \) into the equation \( 3a - b=13 \):
\(3\times\frac{7}{2}-b = 13\)
\(\frac{21}{2}-b = 13\)
\(b=\frac{21}{2}-13=\frac{21 - 26}{2}=-\frac{5}{2}\)

Answer:

\(a=\frac{7}{2}\) and \(b =-\frac{5}{2}\)