Core Ideas in Plain Terms

Study Sheet

Core Ideas in Plain Terms

The bridge to calculus, in everyday language

Functions and Transformations

Concept
A function is a machine

A function takes an input, does something to it, and gives one output. Writing f(x)f(x) just means "the output when the input is xx." Changing the formula slightly moves the graph: f(x)+kf(x)+k slides it up kk, f(xh)f(x-h) slides it right hh, and a minus sign in front flips it over.

Example
Shifting a parabola
xy(2, 3)

y=(x2)2+3y = (x-2)^2 + 3 is the basic y=x2y = x^2 moved 22 to the right and 33 up, so its lowest point is at (2,3)(2, 3).

Exponentials and Logarithms

Concept
Logarithms undo exponents

The equation logbx=y\log_b x = y is just another way of writing by=xb^y = x — it asks "what power turns bb into xx?" So logs are the reverse of exponentials, and that is why they turn multiplication into addition (log(xy)=logx+logy\log(xy) = \log x + \log y) and pull an exponent down front.

Example
Solving 2^x = 8

log28=3\log_2 8 = 3 because 23=82^3 = 8. The log simply reports the exponent.

Sequences, Series, and Limits

Concept
A limit is where a value is heading

A limit asks: as the input gets closer and closer to some number, what does the output approach? You don't have to reach it — just see where it is heading. This idea is the whole foundation of calculus.

Example
Where 1/x heads as x grows
xy

As xx gets huge, 1x\tfrac1x gets tiny — it heads toward 00. We write limx1x=0\lim_{x\to\infty}\tfrac1x = 0.

Going Deeper: Log Rules and Infinite Series

Concept
Log rules are exponent rules in reverse

Multiplying powers adds exponents, so logs turn multiplication into addition — the product rule, log(ab)=loga+logb\log(ab) = \log a + \log b — and pull exponents down front — the power rule, log(an)=nloga\log(a^n) = n\log a. That second rule is why logs solve for unknowns stuck in an exponent — the nn comes down where you can reach it.

Reminder — Logarithm rules:logb(xy)=logbx+logby,logb ⁣xy=logbxlogby,logb(xn)=nlogbx\log_b(xy)=\log_b x+\log_b y,\quad \log_b\!\frac{x}{y}=\log_b x-\log_b y,\quad \log_b(x^{n})=n\log_b x
Reminder — The differentiation rules:(xn)=nxn1,(fg)=fg+fg,(fg)=fgfgg2,(f(g(x)))=f(g(x))g(x)(x^{n})'=nx^{n-1},\quad (fg)'=f'g+fg',\quad \left(\tfrac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}},\quad \big(f(g(x))\big)'=f'(g(x))g'(x)
Concept
How an infinite series can add to a finite number

In a geometric series each term is the same fraction rr of the one before. When r<1|r| < 1 the terms shrink fast enough that the total closes in on the infinite geometric series formula S=a1rS = \dfrac{a}{1 - r}. So 1+12+14+18+=1112=21 + \tfrac12 + \tfrac14 + \tfrac18 + \cdots = \dfrac{1}{1 - \frac12} = 2: you never pass 2, but you get as close as you like.

Example
Putting the log rules to work

log28+log24=3+2=5\log_2 8 + \log_2 4 = 3 + 2 = 5, and indeed log2(84)=log232=5\log_2(8 \cdot 4) = \log_2 32 = 5. Adding the logs multiplied the numbers.

Example
A function and its inverse (multi-step)

Let f(x)=2x3f(x) = 2x - 3.

Step 1 — evaluate: f(5)=7f(5) = 7.

Step 2 — find the inverse by swapping and solving: x=2y3x = 2y - 3 gives f1(x)=x+32f^{-1}(x) = \dfrac{x + 3}{2}.

Step 3 — check with a composition: f1(11)=142=7f^{-1}(11) = \dfrac{14}{2} = 7 and f(7)=11f(7) = 11 — the functions undo each other.

Problem-Solving Playbook

Concept
Rewrite before you grind

Most precalc problems get easy after ONE rewrite: exponential equations → same base; log equations → exponent form (logbx=y    by=x\log_b x = y \iff b^y = x); rational expressions → factor first. Only start computing after the rewrite.

Example
Worked: same-base rewrite

Solve 4x=8x14^x = 8^{x-1}. Both are powers of 22: 22x=23(x1)2^{2x} = 2^{3(x-1)}. Equal bases force equal exponents: 2x=3x32x = 3x - 3, so x=3x = 3. (Check: 43=644^3 = 64 and 82=648^{2} = 64.)