Finding Unknown Parameters from Continuity

Continuity of a function is a fundamental concept in calculus that describes the behavior of functions as they approach certain points. A function is said to be continuous at a point if the limit of the function as it approaches the point is equal to the function's value at that point. When dealing with piecewise functions or functions with parameters, ensuring continuity can involve finding unknown parameters that make the function continuous across its domain.

Understanding Continuity

Before we delve into finding unknown parameters, let's understand what continuity means. A function $f(x)$ is continuous at a point $x = a$ if the following three conditions are met:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists.
3. $\lim_{x \to a} f(x) = f(a)$.

In simpler terms, there should be no breaks, jumps, or holes in the graph of the function at $x = a$.

Continuity in Piecewise Functions

Piecewise functions are functions defined by different expressions over different intervals. Ensuring continuity in such functions often requires finding parameters that will make the function's value the same from both sides of the interval boundary.

Consider a piecewise function defined as follows:

$$ f(x) = \begin{cases} g(x) & \text{for } x < c \ h(x) & \text{for } x \geq c \end{cases} $$

To ensure continuity at $x = c$, we must have:

$$ \lim_{x \to c^-} g(x) = \lim_{x \to c^+} h(x) = f(c) $$

Finding Unknown Parameters

When a function includes unknown parameters, we can often find their values by setting up equations based on the continuity conditions. Let's look at an example to illustrate this process.

Example 1: Linear Piecewise Function

Suppose we have a piecewise function with an unknown parameter $a$:

$$ f(x) = \begin{cases} ax + 2 & \text{for } x < 3 \ 4x - a & \text{for } x \geq 3 \end{cases} $$

We want to find the value of $a$ that makes $f(x)$ continuous at $x = 3$.

Solution:

To ensure continuity at $x = 3$, we must have:

$$ \lim_{x \to 3^-} (ax + 2) = \lim_{x \to 3^+} (4x - a) = f(3) $$

Evaluating the limits, we get:

$$ a(3) + 2 = 4(3) - a $$

Solving for $a$, we find:

$$ 3a + 2 = 12 - a \ 4a = 10 \ a = \frac{10}{4} \ a = 2.5 $$

Thus, $a = 2.5$ makes the function continuous at $x = 3$.

Table of Differences and Important Points

Point of Continuity	Left-Hand Limit	Right-Hand Limit	Value of Function	Condition for Continuity
$x = c$	$\lim_{x \to c^-} g(x)$	$\lim_{x \to c^+} h(x)$	$f(c)$	$\lim_{x \to c^-} g(x) = \lim_{x \to c^+} h(x) = f(c)$

Formulas for Continuity

• Continuity at a Point: $\lim_{x \to a} f(x) = f(a)$
• Continuity of Piecewise Functions: $\lim_{x \to c^-} g(x) = \lim_{x \to c^+} h(x) = f(c)$

Examples to Explain Important Points

Example 2: Quadratic-Cubic Piecewise Function

Consider a piecewise function with two unknown parameters $a$ and $b$:

$$ f(x) = \begin{cases} ax^2 + b & \text{for } x < 1 \ x^3 - 3x + a & \text{for } x \geq 1 \end{cases} $$

Find the values of $a$ and $b$ that make $f(x)$ continuous at $x = 1$.

Solution:

For continuity at $x = 1$, we must have:

$$ \lim_{x \to 1^-} (ax^2 + b) = \lim_{x \to 1^+} (x^3 - 3x + a) = f(1) $$

Evaluating the limits, we get two equations:

$$ a(1)^2 + b = 1^3 - 3(1) + a \ a + b = -2 + a $$

Since $a$ cancels out, we find that $b = -2$. Now, we need to find $a$ by evaluating $f(1)$:

$$ f(1) = 1^3 - 3(1) + a = -2 + a $$

Since $f(1)$ must also equal $a + b$, and we know $b = -2$, we have:

$$ a - 2 = -2 + a $$

This equation is true for any $a$, so $a$ can be any real number. Thus, $b = -2$ and $a$ is any real number to ensure continuity at $x = 1$.

In conclusion, finding unknown parameters from continuity involves setting up equations based on the limit definitions of continuity and solving for the parameters. This process ensures that the function behaves smoothly across its domain without any breaks or jumps.