Understanding Problems Based on Investigating Functions Using Functional Relationship

Investigating functions through their functional relationships involves understanding how functions interact with each other, how they can be combined, and how they can be manipulated to reveal certain properties such as continuity, differentiability, and limits. In this context, we will explore various aspects of functional relationships and how they can be used to solve problems in calculus.

Functional Relationships

A functional relationship is a connection or a rule that defines how one quantity depends on another. In mathematics, this often refers to how the output of a function is determined by its input.

Types of Functional Relationships

• Direct Relationship: When one quantity increases, the other also increases.
• Inverse Relationship: When one quantity increases, the other decreases.
• Composite Functions: Functions composed of other functions, denoted as $(f \circ g)(x) = f(g(x))$.
• Inverse Functions: Functions that reverse the effect of the original function, denoted as $f^{-1}(x)$.

Investigating Continuity and Differentiability

When investigating functions, two of the most important properties to consider are continuity and differentiability.

• Continuity: A function $f(x)$ is continuous at a point $x = a$ if $\lim_{x \to a} f(x) = f(a)$.
• Differentiability: A function $f(x)$ is differentiable at a point $x = a$ if $f'(a)$ exists.

Table of Differences and Important Points

Property	Continuity	Differentiability
Definition	$\lim_{x \to a} f(x) = f(a)$	$f'(a)$ exists
Graphical View	No breaks, jumps, or holes	Smooth curve, no sharp corners
Composite Func.	Continuity of both $f(x)$ and $g(x)$	Differentiability of both $f(x)$ and $g(x)$
Inverse Func.	Continuous if original is continuous	Differentiable if original is differentiable

Investigating Functions Using Functional Relationships

Composite Functions

When dealing with composite functions, we need to investigate the continuity and differentiability of both the inner function $g(x)$ and the outer function $f(x)$.

Continuity of Composite Functions

For $(f \circ g)(x)$ to be continuous at $x = a$, both $f(x)$ and $g(x)$ must be continuous at $g(a)$ and $a$, respectively.

Differentiability of Composite Functions

For $(f \circ g)(x)$ to be differentiable at $x = a$, both $f(x)$ and $g(x)$ must be differentiable at $g(a)$ and $a$, respectively. The derivative is given by the chain rule:

$$ (f \circ g)'(a) = f'(g(a)) \cdot g'(a) $$

Inverse Functions

Inverse functions reverse the effect of the original function. For an inverse function to exist, the original function must be one-to-one (bijective).

Continuity of Inverse Functions

If $f(x)$ is continuous and bijective on an interval, then its inverse $f^{-1}(x)$ is also continuous on the corresponding interval.

Differentiability of Inverse Functions

If $f(x)$ is differentiable and its derivative is never zero on an interval, then $f^{-1}(x)$ is differentiable on the corresponding interval. The derivative of the inverse function is given by:

$$ (f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))} $$

Examples

Example 1: Composite Functions

Let $f(x) = \sqrt{x}$ and $g(x) = x^2 + 1$. Investigate the continuity and differentiability of the composite function $(f \circ g)(x)$.

Solution:

• $f(x)$ is continuous for $x \geq 0$.
• $g(x)$ is continuous for all $x$.
• Therefore, $(f \circ g)(x)$ is continuous for all $x$.
• $f(x)$ is differentiable for $x > 0$.
• $g(x)$ is differentiable for all $x$.
• Therefore, $(f \circ g)(x)$ is differentiable for all $x \neq 0$.

Example 2: Inverse Functions

Let $f(x) = e^x$. Find the derivative of its inverse function.

Solution:

• $f(x)$ is continuous and differentiable for all $x$.
• The inverse function is $f^{-1}(x) = \ln(x)$.
• Using the formula for the derivative of the inverse function:

$$ (f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))} = \frac{1}{e^{\ln(y)}} = \frac{1}{y} $$

Therefore, the derivative of the inverse function $f^{-1}(x) = \ln(x)$ is $\frac{1}{x}$.

By understanding the functional relationships and applying the properties of continuity and differentiability, we can investigate and solve complex problems involving functions. It is crucial to remember the conditions under which these properties hold and to apply the appropriate rules and formulas when analyzing composite and inverse functions.