Mathematics 31

Operations act pointwise: $(f+g)(x)$$=f(x)+g(x)$, and similarly for −, × and ÷ (the quotient excludes zeros of the denominator).

In $y=a\,f[k(x+c)]+d$: a is vertical stretch/reflection, d is vertical shift; k is horizontal stretch/reflection, c is horizontal shift — the horizontal ones act INSIDE and behave inversely.

Factoring extends to integral AND rational exponents; a common technique is factoring out the smallest power.

$\lim_{x \to a} f(x)=L$ means f(x) gets ARBITRARILY CLOSE to L as x approaches a — the value f(a) itself need not equal L (or even exist).

Composition CHAINS functions: $(f\circ g)(x)=f(g(x))$ — do the inside first; the domain must keep g(x) inside f's domain.

A secant through (x, f(x)) and (x + h, f(x + h)) has slope $\frac{f(x+h)-f(x)}{h}$; letting h → 0 sends the secant to the TANGENT.

Product: $(uv)'=u'v+uv'$. Quotient: $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}$. Chain: $\frac{d}{dx}f(g(x))$$=f'(g(x))\,g'(x)$.

Differentiate both sides with respect to x, applying the CHAIN RULE to every y (each y-term gains a $\frac{dy}{dx}$), then solve algebraically for $\frac{dy}{dx}$.

Higher derivatives repeat differentiation: $f''(x)=\frac{d}{dx}f'(x)$, then $f'''(x)$, and so on — each one differentiates the previous result.

$\frac{d}{dx}\sin x=\cos x$, $\frac{d}{dx}\cos x=-\sin x$, $\frac{d}{dx}\tan x=\sec^2 x$, plus the three reciprocal ratios — each combines with the chain, product and quotient rules.

f'(x) > 0 → rising, f'(x) < 0 → falling; critical points occur where f'(x) = 0 or is undefined. f''(x) > 0 → concave up, f''(x) < 0 → concave down.

Write an equation relating the quantities, then differentiate BOTH sides with respect to time; the chain rule attaches a $\frac{d}{dt}$ rate to each changing variable.

A sign CHANGE of f' at a critical point gives a max or min; a sign change of f'' marks an inflection point. Zero derivative alone is only ONE possible condition for an extremum.

Substitute the known instantaneous values only AFTER differentiating — Leibnitz notation ($\frac{dV}{dt}$, $\frac{dr}{dt}$) keeps the rates straight.

Translate the problem into ONE function of one variable using a constraint, differentiate, set f'(x) = 0, and test the critical point to confirm a max or min.

Antidifferentiation reverses the power rule: $\int x^n\,dx$$=\frac{x^{n+1}}{n+1}+C$ (n ≠ −1) — always add the constant C.

Partition [a, b] into n strips and sum rectangle areas; the true area is the LIMIT of these sums as n → ∞ (rectangle width → 0).

The indefinite integral is an antiderivative plus C; the DEFINITE integral $\int_a^b f(x)\,dx$ is the single number $F(b) - F(a)$, with the constant cancelling.

Velocity is the derivative of position, acceleration the derivative of velocity: $v(t)=s'(t)$, $a(t)=v'(t)$. Antidifferentiate to go back: $v=\int a\,dt$, $s=\int v\,dt$.

Many functions share a derivative; they differ only by C, so an antiderivative names a FAMILY of curves. A given point picks out one member.

$\frac{d}{dx}e^x=e^x$ and $\int e^x\,dx=e^x+C$; $\frac{d}{dx}\ln x=\frac{1}{x}$.

$y'=ky$ solves as $y=Ae^{kx}$ — growth for k > 0, decay for k < 0; A is the initial amount.

Newton-Raphson iterates $x_{n+1}$$=x_n-\frac{f(x_n)}{f'(x_n)}$ toward a root — an ITERATIVE procedure that refines a guess by following the tangent line.

Upper and lower Riemann sums bracket the true integral; every quadrature rule INTERPOLATES between them, and more subdivisions raise accuracy.

Let u = the inner function; then $du = u'(x)\,dx$ replaces that factor, turning the integral into one in u. For a DEFINITE integral, change the limits to u-values.

Rotating y = f(x) about the x-axis makes each thin slice a DISC of radius f(x) and area π[f(x)]²; summing them is $V$$=\pi\int_a^b [f(x)]^2\,dx$.

$y''(t) = f(t)$ solves by antidifferentiating twice; $y''(t) = -k^2 y$ is the simple-harmonic equation whose solutions are sinusoids.

$y'=ky$ solves as $y=Ae^{kx}$ (growth k > 0, decay k < 0); $y'=k(y-y_0)$ models return to equilibrium. Verify a proposed solution by substituting it back.

Model revenue, profit or cost as a function of price or quantity, differentiate, and set the derivative to zero to locate the optimum — then confirm max vs min.

For the region between two curves, find the intersection points first; those set the limits (and the washer uses outer² − inner²).

An EXAMPLE supports a claim; a COUNTEREXAMPLE disproves it. Intuitive arguments suggest, rigorous proofs establish — and every theorem has conditions that must hold.

The intermediate value, Rolle's, mean value theorems and the FTC each guarantee an existence; rigorous derivative proofs are built from the corresponding limit theorems.