Question:

Consider the function

1. Determine the values of $a$ and $b$ for which the function $f(x)$ is continuous at $x = 1$.
2. Identify and classify any other types of discontinuities (removable, jump, infinite) for the function $f(x)$.

Answer:

To determine the values of $a$ and $b$ for which the function $f(x)$ is continuous at $x = 1$, we need to check if the following conditions are met:

1. The limit of $f(x)$ as $x$ approaches 1 from the left side is equal to the limiting value as $x$ approaches 1 from the right side.
2. The limiting value as $x$ approaches 1 from either side is equal to the value of $f(x)$ at $x = 1$.

Let's begin by finding the limit of $f(x)$ as $x$ approaches 1 from the left side, denoted as $L_1$:

Substituting the given function:

We can simplify the expression using algebraic manipulation:

Since both the numerator and denominator have common factors of $(x-1)$, we can cancel them:

Now, we can evaluate the limit:

Next, let's find the limit of $f(x)$ as $x$ approaches 1 from the right side, denoted as $L_2$:

Substituting the given function:

Again, we can simplify the expression using algebraic manipulation:

Cancelling the common factors:

Evaluating the limit:

Since both $L_1 = 2$ and $L_2 = 2$, we can conclude that the function $f(x)$ is continuous at $x = 1$ if $a = b = 2$.

Now, let's examine other types of discontinuities for the function $f(x)$.

To determine if there are any removable discontinuities, we need to check if the function can be redefined at certain points to make it continuous. In this case, $f(x)$ is already continuous at $x = 1$, so there are no removable discontinuities.

To identify jump discontinuities, we look for abrupt changes in the value of the function as $x$ approaches a specific point. Since we have already established the continuity at $x = 1$, there are no jump discontinuities for $f(x)$.

Lastly, we check for infinite discontinuities by looking for vertical asymptotes. We observe that the function is a rational expression, and there is a potential vertical asymptote if the denominator becomes zero. In this case, as $x$ approaches $1$, the denominator $x - 1$ becomes zero. Thus, we have a vertical asymptote at $x = 1$.

To summarize, for the function $f(x)$:

• The function is continuous at $x = 1$ if $a = b = 2$.
• There are no removable discontinuities.
• There are no jump discontinuities.
• There is a vertical asymptote at $x = 1$.