Problem 24 - The Calculus with Analytic Geometry

Question

Let the function \(f\) be defined by $$ f(x)= \begin{cases}\frac{g(x)-g(a)}{x-a} & \text { if } x \neq a \\\ g^{\prime}(a) & \text { if } x=a\end{cases} $$ Prove that if \(g^{\prime}(a)\) exists, \(f\) is continuous at \(a\).

Step-by-Step Solution

Answer

Since \( \lim\_{x \to a} f(x) = f(a) = g'(a) \), \( f \) is continuous at \( a \).

Step 1 - Define Continuity

A function is continuous at a point if the limit of the function as it approaches that point from both sides is equal to the function's value at that point. Mathematically, for a function \( f(x) \) to be continuous at \( x = a \), it must follow that: \[ \lim \_{x \to a} f(x) = f(a) \]

Step 2 - Evaluate \( f(a) \)

Based on the given definition of \( f(x) \), at \( x = a \), \[ f(a) = g'(a) \]

Step 3 - Evaluate \( \lim\_{x \to a}f(x) \)

To find \( \lim\_{x \to a}f(x) \) when \( x eq a \), we use the first part of the piecewise function: \[ f(x) = \frac{g(x)-g(a)}{x-a} \] Note that this is the definition of the derivative of \( g(x) \) at \( a \). So, \[ \lim\_{x \to a} \frac{g(x)-g(a)}{x-a} = g'(a) \]

Step 4 - Conclude Continuity

Since\( \Omega lim\_{x \to a} f(x) = g'(a) \) by Step 3, and \( f(a) = g'(a) \) by Step 2, we have: \[ \lim \_{x \to a} f(x) = f(a) \] Hence, \( f(x) \) is continuous at \( x = a \).

Key Concepts

Piecewise Functions

Piecewise functions are those that have different definitions depending on the input value. They allow us to model situations where a rule or behavior changes based on certain conditions. For example, in our exercise, the function \( f(x) \) is defined differently for when \( x = a \) and when \( x eq a \). This makes it a piecewise function.

Since piecewise functions can have abrupt changes, continuity (smoothness of the graph) needs to be checked carefully. Each piece needs to align properly at the edges, especially at points where the definition changes.

Understanding how to check continuity involves learning about the limits of the function, which brings us to our next key concept.

Limits

The concept of limits is crucial in calculus and especially important for ensuring continuity at a point. A limit helps us understand the behavior of a function as it approaches a particular point. For a function \( f(x) \) to be continuous at \( x = a \), the limit of \( f(x) \) as \( x \to a \) must be equal to \( f(a) \). Mathematically, this is expressed as:

\[ \lim_{{x \to a}} f(x) = f(a) \]

In our exercise, the limit of \( f(x) \) as \( x \to a \) was calculated using the derivative of \( g(x) \). This fits the structure of our piecewise function. By evaluating the limit using the piece of the function defined for \( x eq a \), we established that:

\[ \lim_{{x \to a}} \frac{{g(x) - g(a)}}{{x - a}} = g'(a) \]

Aligning the limit with the exact value the piecewise function takes at \( x = a \), confirms that \( f(x) \) is indeed continuous at \( a \).

Derivatives

Derivatives represent the rate of change of a function with respect to its variable. They are central in calculus and help us understand how a function changes at any given point. In our problem, the derivative of \( g(x) \) at \( a \) is denoted as \( g'(a) \).

In the context of the original exercise, we used derivatives to determine continuity at the point \( x = a \). Inside the piecewise function, the expression:

\[ \frac{{g(x) - g(a)}}{{x - a}} \]

is recognized as the definition of a derivative. So, as \( x \) approaches \( a \), the expression approaches \( g'(a) \). This rate of change helps confirm that the limit of our function \( f(x) \) aligns perfectly with the value at \( x = a \).

By comprehending rates of change through derivatives, we establish the smooth transition between pieces in a piecewise function, ensuring its continuity.