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limx→a f(x) = L means f(x) gets arbitrarily close to L as x approaches a — from both sides. The value f(a) itself does not have to exist for the limit to exist.
lim f(x) = L ⇔ limx→a⁻ f(x) = L AND limx→a⁺ f(x) = L| If you see… | Try first | If that fails |
|---|---|---|
| plug-in works | direct substitution | — |
| 0/0 | factor & cancel | conjugate / L'Hôpital |
| ∞/∞ | divide by highest power | L'Hôpital |
| √ in numerator | multiply by conjugate | — |
| trig at 0 | sin x / x → 1, (1−cos x)/x → 0 | squeeze theorem |
L'Hôpital (only for 0/0 or ∞/∞): lim f/g = lim f' / g'IVT: if f is continuous on [a,b] and N is between f(a) and f(b), some c ∈ [a,b] has f(c) = N. Use to prove a root exists.
A limit can exist where f isn't even defined (removable hole). Conversely, f(a) can be defined but the limit DNE. Limit and value are independent quantities — read the problem twice.
| Derivative | Sign + | Sign − | = 0 means |
|---|---|---|---|
| f'(x) | increasing | decreasing | critical point (max / min / saddle) |
| f''(x) | concave up ⌣ | concave down ⌢ | inflection candidate |
1. Domain. Vertical asymptotes where f undefined; horizontal as x → ±∞.
2. Find f'. Critical points where f' = 0 or DNE.
3. Sign chart of f' → intervals of increase / decrease.
4. Find f''. Sign chart → concavity + inflection points.
5. Plot intercepts, asymptotes, critical pts, inflection pts. Connect smoothly.
1st derivative test: f'(c) changes + → − ⇒ local MAX | − → + ⇒ local MIN2nd derivative test: f'(c)=0 AND f''(c) > 0 ⇒ local MIN | f''(c) < 0 ⇒ local MAXAsymptotes: vertical at x = a if limx→a f = ±∞. Horizontal y = L if limx→±∞ f = L. Slant when degree(num) = degree(den) + 1.
f'(c) = 0 only tells you c is a candidate. f(x) = x³ has f'(0) = 0 but no extremum. Always sign-check f' on both sides before declaring max/min.
f'(x) = limh→0 [f(x+h) − f(x)] / hThe derivative is the instantaneous rate of change — the slope of the tangent line at x. Geometrically, the secant slope between (x, f(x)) and (x+h, f(x+h)) becomes the tangent slope as h → 0.
| Rule | Form | Result |
|---|---|---|
| Constant | d/dx[c] | 0 |
| Power | d/dx[xⁿ] | n·x^(n−1) |
| Sum | d/dx[f ± g] | f' ± g' |
| Product | d/dx[f·g] | f'·g + f·g' |
| Quotient | d/dx[f/g] | (f'·g − f·g') / g² |
| Chain | d/dx[f(g(x))] | f'(g(x))·g'(x) |
(sin x)' = cos x (cos x)' = −sin x (eˣ)' = eˣ (ln x)' = 1/xNotation: f'(x) = dy/dx = Dxf. They mean the same thing; don't let multiple notations rattle you on the exam.
Students see f·g and write f'·g'. Wrong. The derivative of a product is not the product of derivatives. It's f'g + fg'. This single mistake costs more partial-credit points than any other in Calc 1.
1. Verify f continuous on [a, b].
2. Find critical points c in (a, b) where f'(c) = 0 or DNE.
3. Evaluate f at every critical point AND at the endpoints a, b.
4. Largest = absolute MAX. Smallest = absolute MIN.
Extreme Value Theorem: f continuous on [a, b] ⇒ f attains absolute max and min| Step | What you write |
|---|---|
| 1 | Define variables + draw a picture |
| 2 | Write the objective Q to optimize |
| 3 | Write the constraint equation |
| 4 | Use constraint to reduce Q to ONE variable |
| 5 | Differentiate, set Q' = 0, solve |
| 6 | Verify max vs min (1st or 2nd test) |
Optimization without endpoints (open interval / all of ℝ): you must justify max/min by 2nd derivative or sign analysis — EVT does not apply.
On a closed interval, the absolute max can occur at an endpoint, not at any critical point. Students compute critical points, pick the larger one, and miss that f(a) or f(b) was bigger. Always evaluate f at a, b, AND every critical pt.
F is an antiderivative of f if F'(x) = f(x). Antiderivatives differ only by a constant — that's why +C exists.
∫ xⁿ dx = x^(n+1)/(n+1) + C (n ≠ −1)
∫ 1/x dx = ln|x| + C ∫ eˣ dx = eˣ + C
∫ sin x dx = −cos x + C ∫ cos x dx = sin x + C∫ab f(x) dx = limn→∞ Σi=1n f(xᵢ*) Δx, Δx = (b−a)/nThe definite integral is the signed area between f(x) and the x-axis. Above the axis = +, below = −.
| Sum type | xᵢ* | For increasing f |
|---|---|---|
| Left | xᵢ₋₁ | underestimate |
| Right | xᵢ | overestimate |
| Midpoint | (xᵢ₋₁ + xᵢ)/2 | usually best |
d/dx ∫ax f(t) dt = f(x)Differentiation undoes integration.
∫ab f(x) dx = F(b) − F(a)Where F is any antiderivative of f.
U-substitution: when integrand is f(g(x))·g'(x), let u = g(x), du = g'(x) dx. Reduces to ∫ f(u) du. The single most powerful Calc 1 integration trick.
Indefinite integrals always need +C. Definite integrals don't. And ∫ba f = −∫ab f — flipping the limits flips the sign. Sloppy bookkeeping = lost points.
d/dx [ f(g(x)) ] = f'(g(x)) · g'(x)Differentiate the outer function (treating the inner as a single variable), then multiply by the derivative of the inner. Repeat for nested compositions.
| Function | Derivative |
|---|---|
| (3x + 5)⁷ | 7(3x + 5)⁶ · 3 = 21(3x + 5)⁶ |
| sin(x²) | cos(x²) · 2x |
| e^(x³) | e^(x³) · 3x² |
| ln(cos x) | (1/cos x) · (−sin x) = −tan x |
When y is defined implicitly by an equation in x and y, differentiate both sides w.r.t. x, treating y as a function of x. Every y term gets a dy/dx multiplier via the chain rule. Then solve for dy/dx.
x² + y² = 25 ⇒ 2x + 2y · dy/dx = 0 ⇒ dy/dx = −x / y1. Draw + label variables (use letters, not numbers yet).
2. Write the geometric / physical relation tying variables together.
3. Differentiate both sides w.r.t. t (every variable gets a d/dt).
4. Now plug in the instantaneous values.
5. Solve for the unknown rate.
d/dx[sin(2x)] is not cos(2x). It's 2 cos(2x). The single most common chain-rule oversight: doing the outer derivative and forgetting to multiply by the derivative of the inner function. Cost: full credit on the problem.
y − f(a) = f'(a) · (x − a)Slope = f'(a) (the derivative evaluated at the point). Passes through (a, f(a)). Same point-slope form you learned in Algebra 2 — but with a derivative, not a numeric slope.
L(x) = f(a) + f'(a)(x − a) ← used for tiny changes near a
dy = f'(x) dx ← differential formLinearization replaces f near a with its tangent line. Used to approximate values like √4.05 ≈ 2 + (1/4)(0.05) = 2.0125 (true value: 2.01246).
If f is continuous on [a, b] AND differentiable on (a, b),
then ∃ c ∈ (a, b) with f'(c) = (f(b) − f(a)) / (b − a).
Translation: somewhere on the interval, the instantaneous slope equals the average slope.
Linear approximation error: for a smooth function, error scales like (Δx)². Doubling Δx roughly quadruples error.
Find the tangent to f(x) at the point where x = 2: students compute f'(x) and use it as the slope. You need f'(2), the slope evaluated at x = 2, plus f(2) for the y-coordinate. Always plug in.
| Keyword / setup | Use § from | Tool / move |
|---|---|---|
| 'limit' or 'find lim' | § ① | direct sub → factor → conjugate → L'Hôpital |
| 0/0 or ∞/∞ | § ① | L'Hôpital (only these forms!) |
| 'continuous at' / 'removable' | § ① | 3-step continuity check |
| 'find f'(x)' / 'differentiate' | § ② | power · product · quotient · chain rule |
| composite f(g(x)) | § ③ | chain rule: outer'·inner' |
| equation in x AND y | § ③ | implicit diff (every y → dy/dx) |
| 'how fast' / two variables changing | § ③ | related rates (5-step recipe) |
| 'tangent line at x = a' | § ④ | y − f(a) = f'(a)(x − a) |
| 'approximate √…' near nice number | § ④ | linearization L(x) |
| 'show ∃ c with …' on [a,b] | § ④ | MVT or Rolle's |
| 'maximum / minimum on [a, b]' | § ⑤ | closed-interval method |
| word problem: 'largest box / least cost' | § ⑤ | objective + constraint, reduce to 1 var |
| 'sketch the graph of f' | § ⑥ | 5-step sketch (domain → f' → f'') |
| 'concave up/down' | § ⑥ | sign of f'' |
| 'inflection point' | § ⑥ | f'' sign change |
| 'find ∫…dx' (no limits) | § ⑦ | indefinite — don't forget +C |
| '∫ from a to b' / 'area under' | § ⑦ | FTC: F(b) − F(a) |
| integrand has f(g)·g' | § ⑦ | u-sub (change limits!) |
| 'd/dx ∫ from a to x f(t) dt' | § ⑦ | FTC Part 1: just f(x) |
Half of partial-credit losses come from picking the wrong tool. Re-read the question, identify the keyword, then map to a section above. The model selection IS the problem-solving step.
Mixing dy/dx and y' is fine, but be consistent within one problem. Mark every dy/dx in implicit diff. Show u, du explicitly in u-sub. Sloppy notation = lost points even when math is right.