for what values of a and b is the following function continuous at every x? f(x)=\\begin{cases}-1&x\\leq…

for what values of a and b is the following function continuous at every x? f(x)=\\begin{cases}-1&x\\leq - 1\\ax - b&-1<x<3\\13&x\\geq3\\end{cases} find the values of a and b for which the function f continuous at every x. a = and b = (type integers or simplified fractions.)

for what values of a and b is the following function continuous at every x? f(x)=\\begin{cases}-1&x\\leq - 1\\ax - b&-1<x<3\\13&x\\geq3\\end{cases} find the values of a and b for which the function f continuous at every x. a = and b = (type integers or simplified fractions.)

Answer

Explanation:

Step1: Use left - hand limit at $x=-1$

For $x\leq - 1$, $f(x)=-1$. For $-1 < x < 3$, $f(x)=ax - b$. The left - hand limit as $x\to - 1$ of $ax - b$ is $-a - b$, and it should equal $-1$. So, $-a - b=-1$ or $a + b = 1$.

Step2: Use right - hand limit at $x = 3$

For $-1 < x < 3$, $f(x)=ax - b$. For $x\geq3$, $f(x)=13$. The right - hand limit as $x\to3$ of $ax - b$ is $3a - b$, and it should equal $13$. So, $3a - b=13$.

Step3: Solve the system of equations

We have the system of equations $\begin{cases}a + b=1\3a - b=13\end{cases}$. Add the two equations: $(a + b)+(3a - b)=1 + 13$. Simplifying gives $4a=14$, so $a=\frac{7}{2}$.

Step4: Find the value of $b$

Substitute $a = \frac{7}{2}$ into $a + b=1$. Then $\frac{7}{2}+b=1$, and $b=1-\frac{7}{2}=-\frac{5}{2}$.

Answer:

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