Welcome back to class! In our session on September 7th, 2023, we focused heavily on the "Algebra of Functions." We moved beyond simple evaluation into how functions interact with one another through arithmetic operations, composition, and inversion. This foundational knowledge is essential for calculus, particularly when we reached the difference quotient.
1. Operations with Functions
Just like numbers, functions can be added, subtracted, multiplied, and divided. We looked at combining polynomials and rational functions. The key takeaway here is always minding the domain.
- Sum/Difference: $(f \pm g)(x) = f(x) \pm g(x)$
- Product: $(f \cdot g)(x) = f(x) \cdot g(x)$
- Quotient: $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$, provided $g(x) \neq 0$.
We practiced simplifying complex rational expressions, such as $\frac{\frac{2}{x+8}}{\frac{9}{x}}$, where we multiply by the reciprocal to simplify the expression to $\frac{2x}{9x+72}$, while noting that $x \neq -8$ and $x \neq 0$.
2. The Difference Quotient
One of the most important bridges to calculus is the Difference Quotient. We worked through a quadratic example: $$f(x) = -4x^2 - 6x + 6$$
To find $\frac{f(x+h)-f(x)}{h}$, we follow three strict steps:
- Find $f(x+h)$: Substitute $(x+h)$ into every instance of $x$.
Result: $-4(x^2+2xh+h^2) - 6(x+h) + 6$ - Subtract $f(x)$: Notice how the original terms ($-4x^2$, $-6x$, $+6$) cancel out.
Result: $-8xh - 4h^2 - 6h$ - Divide by $h$: Factor an $h$ out of the numerator and cancel it with the denominator.
Final Answer: $-8x - 4h - 6$
3. Composition of Functions
Composition involves plugging one function inside another, denoted as $(f \circ g)(x) = f(g(x))$. We looked at this numerically, algebraically, and using mapping diagrams.
For example, given $q(x) = x^2 + 4$ and $r(x) = \sqrt{x+5}$, the composition $(q \circ r)(x)$ simplifies nicely because the square root and the square cancel out:
$$q(\sqrt{x+5}) = (\sqrt{x+5})^2 + 4 = (x+5) + 4 = x+9$$Note: Always check the domain of the inner function first!
4. Inverse Functions
Finally, we discussed how to undo a function. A function has an inverse if it passes the Horizontal Line Test (is one-to-one). We solved for inverses algebraically by swapping $x$ and $y$ and solving for the new $y$.
A challenging example from class involved the rational function $h(x) = \frac{9x-8}{7x+4}$. To find the inverse:
- Set $y = \frac{9x-8}{7x+4}$
- Swap variables: $x = \frac{9y-8}{7y+4}$
- Multiply and group $y$ terms: $x(7y+4) = 9y-8 \implies 7xy + 4x = 9y - 8$
- Factor out $y$ and divide: $y(7x-9) = -4x-8$
- Result: $h^{-1}(x) = \frac{-4x-8}{7x-9}$
Keep practicing these algebraic manipulations—they are the toolkit you will use for the rest of the semester!