Calculus

Calculus studies local change and accumulated quantity. Before applying a theorem, check its hypotheses: continuity, differentiability, interval endpoints, and whether a denominator can vanish. For optimization, locate candidates and then compare values or use a justified sign test. For integration, choose a substitution or symmetry that matches the structure.

Limits, continuity, and differentiability

Idea. A limit describes nearby values, not necessarily the function value at the point. A function is continuous at a if its value exists and equals the two-sided limit. Differentiability at an interior point implies continuity there, but continuity alone does not guarantee differentiability.

Solved examples.

1. Question. Evaluate $\lim_{x\to2}(3x+1)$. Solution. (1) Polynomials are continuous everywhere. (2) Substitute x=2: $3(2)+1=7$. Answer. 7. Shortcut. Direct substitution works for continuous expressions.
2. Question. Evaluate $\lim_{x\to1}(x^2-1)/(x-1)$. Solution. (1) Direct substitution gives 0/0, so factor. (2) $x^2-1=(x-1)(x+1)$. (3) For x≠1, quotient is x+1. (4) Limit is 2. Answer. 2. Trap. Cancellation simplifies a punctured neighborhood; it does not redefine the original at x=1.
3. Question. Let f(x)=x² for x≠1 and f(1)=5. Is f continuous at 1? Solution. (1) Limit of x² at 1 is 1. (2) Function value is 5. (3) They differ. Answer. Not continuous. Trap. A finite limit alone is not enough; compare the assigned value.
4. Question. Is $f(x)=|x|$ differentiable at zero? Solution. (1) Left derivative is -1. (2) Right derivative is 1. (3) One-sided derivatives differ. Answer. No. It is continuous at zero but not differentiable. Useful fact. A sharp corner often signals unequal one-sided derivatives.
5. Question. Evaluate $\lim_{x\to0}\sin(5x)/x$. Solution. (1) Rewrite as $5\,[\sin(5x)/(5x)]$. (2) Use $\lim_{u\to0}\sin u/u=1$. (3) Result is 5. Answer. 5. Condition. Angles are measured in radians.

Maxima and minima

Idea. An interior differentiable local extremum must have derivative zero, but a stationary point need not be an extremum. On a closed bounded interval, a continuous function attains absolute maximum and minimum. Compare all critical points and endpoints; the first-derivative sign change gives a local classification.

Solved examples.

1. Question. Find local extrema of $f(x)=x^2$. Solution. (1) $f'(x)=2x$. (2) Critical point x=0. (3) Derivative changes from negative to positive. Answer. Local and absolute minimum 0 at x=0; no maximum on $\mathbb R$. Trap. On an unbounded domain, a maximum need not exist.
2. Question. Find maximum and minimum of $f(x)=x^2-4x$ on [0,5]. Solution. (1) $f'=2x-4=0$ at x=2. (2) Evaluate endpoints and critical point: f(0)=0, f(2)=-4, f(5)=5. Answer. Minimum -4 at 2; maximum 5 at 5. Shortcut. Closed interval requires checking endpoints.
3. Question. Classify x=0 for $f(x)=x^3$. Solution. (1) Derivative is $3x^2$, zero at 0. (2) Function increases on both sides. (3) Values pass through the point without turning. Answer. Neither a max nor a min; it is a stationary inflection point. Trap. $f'=0$ is only a candidate condition.
4. Question. A rectangle has perimeter 20. Find dimensions for greatest area. Solution. (1) Let sides x and 10-x. (2) Area $A=x(10-x)=10x-x^2$, for $0\le x\le10$. (3) $A'=10-2x=0$ at 5; the quadratic opens downward. Answer. 5 by 5, area 25. Trap. State the feasible interval before optimizing.
5. Question. Find the minimum of $f(x)=x+9/x$ for x>0. Solution. (1) $f'=1-9/x^2$. (2) Critical point x=3. (3) For x<3 derivative is negative; for x>3 positive. (4) Boundary behavior tends to infinity at 0+ and infinity. Answer. Minimum 6 at x=3. Shortcut. Positive reciprocal terms often admit AM-GM, but derivative confirms the domain-wide result.

Mean value theorem

Idea. If f is continuous on [a,b] and differentiable on (a,b), some c in (a,b) satisfies $f'(c)=[f(b)-f(a)]/(b-a)$. Rolle's theorem is the special case f(a)=f(b), giving some c with f'(c)=0. Verify both regularity conditions.