Problems Based on Finding Unknown Parameters

In mathematics, particularly in calculus, problems involving finding unknown parameters often arise when dealing with functions that are required to be continuous or differentiable. These parameters could be constants in a function or coefficients in a polynomial. To ensure continuity or differentiability, these parameters must satisfy certain conditions.

Continuity

A function $f(x)$ is said to be 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. $\lim_{x \to a} f(x) = f(a)$.

For a function to be continuous on an interval, it must be continuous at every point in that interval.

Differentiability

A function $f(x)$ is differentiable at a point $x = a$ if the derivative $f'(a)$ exists. This implies that the function is smooth (without any sharp corners or cusps) at that point. If a function is differentiable at a point, it is also continuous at that point, but the converse is not necessarily true.

Finding Unknown Parameters

When we have a function with unknown parameters, we often need to find the values of these parameters that make the function continuous or differentiable at a given point or over an interval.

Table of Differences and Important Points

Aspect	Continuity	Differentiability
Definition	No sudden jumps or breaks in the graph	The graph has a well-defined tangent at each point
Mathematical Condition	$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$	$f'(a)$ exists and is finite
Graphical Interpretation	The graph can be drawn without lifting the pen	The graph has a smooth curve without sharp corners
Implication	Differentiability implies continuity	Continuity does not imply differentiability

Formulas

For continuity at $x = a$: $$ \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) $$

For differentiability at $x = a$: $$ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} $$

Examples

Example 1: Ensuring Continuity

Find the value of the unknown parameter $k$ that makes the function $f(x)$ continuous at $x = 3$.

$$ f(x) = \begin{cases} 2x + k & \text{if } x < 3 \ 5 & \text{if } x = 3 \ x^2 - 4x + k & \text{if } x > 3 \end{cases} $$

Solution:

For $f(x)$ to be continuous at $x = 3$, we must have:

$$ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3) $$

Evaluating the left-hand limit:

$$ \lim_{x \to 3^-} (2x + k) = 2 \cdot 3 + k = 6 + k $$

Evaluating the right-hand limit:

$$ \lim_{x \to 3^+} (x^2 - 4x + k) = 3^2 - 4 \cdot 3 + k = 9 - 12 + k = -3 + k $$

Setting the left-hand limit equal to the value of the function at $x = 3$:

$$ 6 + k = 5 $$

Solving for $k$:

$$ k = 5 - 6 = -1 $$

Therefore, $k = -1$ makes the function continuous at $x = 3$.

Example 2: Ensuring Differentiability

Find the value of the unknown parameter $a$ that makes the function $g(x)$ differentiable at $x = 0$.

$$ g(x) = \begin{cases} ax^2 & \text{if } x \leq 0 \ x^3 + 3x^2 & \text{if } x > 0 \end{cases} $$

Solution:

For $g(x)$ to be differentiable at $x = 0$, it must also be continuous at $x = 0$. Therefore, we first ensure continuity:

$$ \lim_{x \to 0^-} ax^2 = a \cdot 0^2 = 0 $$

$$ \lim_{x \to 0^+} (x^3 + 3x^2) = 0^3 + 3 \cdot 0^2 = 0 $$

The function is continuous at $x = 0$ for any value of $a$. Now, we check for differentiability by equating the derivatives from the left and the right at $x = 0$:

$$ \frac{d}{dx}(ax^2) \bigg|{x = 0} = 2ax \bigg|{x = 0} = 0 $$

$$ \frac{d}{dx}(x^3 + 3x^2) \bigg|{x = 0} = (3x^2 + 6x) \bigg|{x = 0} = 0 $$

Since the derivatives from both sides are equal at $x = 0$, the function is differentiable at $x = 0$ for any value of $a$. Therefore, any real number $a$ will make the function $g(x)$ differentiable at $x = 0$.

These examples illustrate how to approach problems involving finding unknown parameters to ensure continuity and differentiability. The key is to apply the definitions and conditions for continuity and differentiability and solve for the parameters accordingly.