Differential Calculus
Differential calculus in Bagrut tests the derivative as a limit and as a tool: tangent lines, extrema, monotonicity, concavity, optimization, and graph analysis. You are expected to justify whether a critical point is a maximum or minimum, not just solve f′(x)=0.
Core rules
- Definition: f′(x)=lim_{h→0} [f(x+h)−f(x)]/h.
- Power rule: (x^n)′=n x^(n−1). Product: (uv)′=u′v+uv′. Quotient: (u/v)′=(u′v−uv′)/v². Chain: [f(g(x))]′=f′(g(x))g′(x).
- Tangent at x=a: y−f(a)=f′(a)(x−a).
- Critical points occur where f′(x)=0 or f′ is undefined. A sign change of f′ determines a local extremum.
- f″>0 means concave up; f″<0 means concave down.
- On a closed interval, compare critical values with endpoint values for global extrema.
Worked example
For f(x)=x³−3x²+2, find the extrema. f′(x)=3x²−6x=3x(x−2), so critical points are x=0 and x=2. f″(x)=6x−6. At x=0, f″=−6<0, so f has a local maximum f(0)=2. At x=2, f″=6>0, so f has a local minimum f(2)=8−12+2=−2. The second derivative test confirms the nature without a sign table.
Traps
Forgetting the chain rule in a composite function is the most frequent error. f′(a)=0 does not by itself prove an extremum; x³ at 0 is a counterexample. Do not ignore endpoints when the question asks for absolute extrema. A tangent line needs both a point and a slope.