Notes on Intervals, Monotonicity, Natural Domain, Binomial Theorem, and f(a+h) Expansions
Interval Types
We focus on an interval I inside the real line; typically an interval is denoted by its endpoints a and b with different bracket types indicating whether endpoints are included.
Open interval:
meaning the set {x:\ a < x < b}; endpoints a and b are not in the interval.
Closed interval:
meaning the set ; endpoints a and b are included in the interval.
Half-open (or half-closed) intervals:
contains a but not b
contains b but not a
Infinite intervals (Infinity is not a number, so we use parentheses around the infinite endpoint):
represents all real x with x > a
represents all real x with x < b
In practice we often consider intervals of the form , , or a finite closed/open mix as above.
Pick two numbers a and b to form an interval; call it I for reference.
Monotonicity: Increasing and Decreasing on an Interval
Definition (in two-point form):
If for all pairs with x1 < x2 in I we have f(x1) < f(x2), then is increasing on I.
If for all pairs with x1 < x2 in I we have f(x1) > f(x2), then is decreasing on I.
Important: To talk about monotonicity, you need at least two points in the interval; you can't talk about a single point.
Quick graph intuition: moving to the right along the x-axis, if the y-values rise, the function is increasing on that stretch; if they fall, it is decreasing on that stretch.
Quick example descriptions:
A graph that keeps rising as x moves to the right is increasing on that interval.
A graph that falls as x increases is decreasing on that interval.
A function that goes up on part of the interval and down on another part is neither increasing nor decreasing on the entire interval.
Tangent-line intuition (preview for tangent-line problems):
If the tangent slope is positive for every point of I, the function is increasing on I.
If the tangent slope is negative for every point of I, the function is decreasing on I.
If the tangent slope changes sign (e.g., positive on one subinterval, negative on another), the function is not increasing or decreasing on the entire interval; there may be a horizontal tangent point that splits I into a left subinterval where it increases and a right subinterval where it decreases.
Natural Domain of a Formula: Example with a Rational Function
Definition: The natural domain of a function given by a formula is the set of inputs x for which the formula makes sense (typically real-valued outputs).
We usually restrict to real numbers unless stated otherwise.
Example:
Consider
Denominator is zero when ; these inputs are not allowed.
Therefore the domain is all real numbers except .
Interval notation form:
Alternative form (explicit):
Another related example (domain with cancellation):
Suppose we have a simplified form after factoring, e.g. something like .
The original domain excludes the point where the denominator is zero, i.e. , even though for all other t the expression equals 1.
Thus the domain is , and the simplified expression is valid only for n\ge 0(a+b)^n = \displaystyle\sum_{k=0}^{n} {n \choose k} a^{n-k} b^{k}n {n \choose k} = \frac{n!}{k!(n-k)!} (a+b)^n (a+b)^5 (a+b)^5 = {5 \choose 0} a^5 + {5 \choose 1} a^4 b + {5 \choose 2} a^3 b^2 + {5 \choose 3} a^2 b^3 + {5 \choose 4} a b^4 + {5 \choose 5} b^5 a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5
In general, the coefficient pattern along row n is: 1, n, (\dfrac{n(n-1)}{2}), …, {n \choose k}, …, 1.
Alternative viewpoint (combinatorial intuition): ${n \choose k}$ counts the number of ways to pick k factors of b from n factors of (a+b).
Practical note on exponents: in each term the exponents satisfy (n-k) + k = n f(x) = x^3f(a+h) = (a+h)^3 = a^3 + 3 a^2 h + 3 a h^2 + h^3.f(a) = a^3f(a+h) - f(a) = (a^3 + 3 a^2 h + 3 a h^2 + h^3) - a^3 = 3 a^2 h + 3 a h^2 + h^3.f(a+h) - f(a) = h \big( 3 a^2 + 3 a h + h^2 \big).h \neq 0\frac{f(a+h) - f(a)}{h} = 3 a^2 + 3 a h + h^2.h\to 0f'(a) = 3 a^2f(x) = 4 + 3x - x^2x=3f(3) = 4 + 3\cdot 3 - 3^2 = 4 + 9 - 9 = 4.f(3+h) = 4 + 3(3+h) - (3+h)^2 = 4 + 9 + 3h - (9 + 6h + h^2) = 4 - 3h - h^2.\frac{f(3+h) - f(3)}{h} = \frac{ (4 - 3h - h^2) - 4 }{h} = \frac{-3h - h^2}{h} = -3 - h, \quad h \neq 0.
This illustrates cancellation and simplification of the quotient to a simple linear expression in h (which would tend to -3 as h -> 0).
Important caution about expansions:
F(a+h) is not equal to F(a) + h. Expanding F(a+h) requires applying the binomial expansion to the term(s) involving h, not simply replacing h by 1 or adding h to F(a).
Always compute the expansion correctly (e.g., for a power n, use (a+h)^n = \sum_{k=0}^n {n \choose k} a^{n-k} h^{k} \frac{f(a+h)-f(a)}{h}h\to 0f'(a){x:\ x\neq -3, x\neq 3}\frac{x+4}{x^2-9}(-\infty,-3)\cup(-3,3)\cup(3,\infty) {n \choose k} = \frac{n!}{k!(n-k)!} (a,b) [a,b] [a,b), (a,b] (a,\infty), (-\infty,b) x1 < x2 \Rightarrow f(x1) < f(x2) x1 < x2 \Rightarrow f(x1) > f(x2) f(x) = \frac{x+4}{x^2-9} \Rightarrow x \neq \pm 3 (-\infty,-3) \cup (-3,3) \cup (3,\infty) \frac{t-2}{t-2} t = 2 (-\infty, 2) \cup (2, \infty) (a+b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^{k} {n \choose k} = \dfrac{n!}{k! (n-k)!} (a+b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5 f(a+h) = (a+h)^3 = a^3 + 3 a^2 h + 3 a h^2 + h^3 f(a+h) - f(a) = 3 a^2 h + 3 a h^2 + h^3 = h(3 a^2 + 3 a h + h^2) h \neq 0 \frac{f(a+h) - f(a)}{h} = 3 a^2 + 3 a h + h^2 f(x) = 4 + 3x - x^2 f(3) = 4 f(3+h) = 4 - 3h - h^2 \frac{f(3+h) - f(3)}{h} = -3 - h $$
These notes connect to broader ideas: monotonicity and derivative intuition, domain restrictions, and efficient algebraic tools (binomial theorem and Pascal’s triangle) that will recur in tangent-line and derivative problems.