Units 4 and 5 apply derivatives to find extreme values, sketch curves, solve optimisation problems, and establish the theoretical foundations of the Mean Value Theorem.
Extreme Value Theorem: continuous function on [a,b] has absolute max and min. Critical points: where f\'(x) = 0 or undefined. Candidates for extrema: critical points + endpoints. Mean Value Theorem: if f is continuous on [a,b] and differentiable on (a,b), then there exists c with f\'(c) = [f(b)-f(a)]/(b-a). Rolle\'s theorem: special case where f(a) = f(b), so f\'(c) = 0.
First derivative test: f\' changes + to - → local max; - to + → local min. Second derivative test: f\'(c)=0 and f\'\'(c)<0 → max; f\'\'(c)>0 → min. Concavity: f\'\' > 0 → concave up (cup); f\'\' < 0 → concave down. Inflection point: concavity changes (f\'\' changes sign). Curve sketching: find domain, intercepts, asymptotes, critical points, intervals of increase/decrease, concavity, inflection points.
Steps: (1) draw diagram, (2) identify quantity to optimise, (3) express as function of one variable using constraint, (4) find critical points, (5) verify max/min (endpoint check or second derivative). L\'Hôpital\'s rule (BC): if lim gives 0/0 or ∞/∞, then lim f(x)/g(x) = lim f\'(x)/g\'(x). May need to apply multiple times.
The Mean Value Theorem (MVT) appears frequently in both multiple choice and free response. Common applications: (1) Given a table of values, justify that a function must have a certain derivative value at some point. (2) Proving that a function must achieve a specific rate of change. (3) Connecting average rate of change to instantaneous rate. Key: always verify the hypotheses (continuous on closed interval, differentiable on open interval) before applying. The exam tests conceptual understanding — "there exists at least one c where the instantaneous rate equals the average rate" — rather than finding c explicitly.
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