Revise: Applications of Derivatives
Critical points, maxima and minima, the second derivative test, and linear approximation.
Differentiation is linear — differentiate each term separately, then combine.
d/dx[3x²−eˣ] = 6x−eˣ.
Critical points: where f'(x)=0 or f'(x) is undefined.
f(x)=x³−3x → critical points at x=±1.
f'(x)>0 ⟹ increasing; f'(x)<0 ⟹ decreasing.
Test one point per interval between critical points.
On [a,b], check critical points AND both endpoints.
f(x)=x³−3x on [−2,2] → abs max 2, abs min −2.
f''(x)>0 ⟹ concave up; f''(x)<0 ⟹ concave down.
f(x)=x³−3x → inflection point at (0,0).
At critical point c: f''(c)>0 ⟹ local min; f''(c)<0 ⟹ local max.
f''(1)=6>0 → local min at x=1.
L(x) = f(a) + f'(a)(x−a) estimates f near a.
√4.1 ≈ 2 + 0.25(0.1) = 2.025.
d/dx[bˣ]=bˣ·ln(b) for general base b, not just eˣ.
d/dx[3ˣ] = 3ˣ·ln3.
Always substitute the critical x-value back into f, not f', for its y-coordinate.
f(−1)=(−1)³−3(−1)=2.
f' is continuous, so it can only change sign at a critical point.
One sample point per interval determines the whole interval.
The Extreme Value Theorem guarantees absolute extrema exist on a closed interval.
Compare all critical-point and endpoint values; largest/smallest wins.
A candidate inflection point (f''=0) is only real if the sign of f'' actually changes there.
f''(x)=2 everywhere → no inflection point at all.
f''(c)=0 is inconclusive — fall back to the First Derivative Test.
Usually faster than checking sign changes on both sides.
Choose a to be a nearby point where f(a) and f'(a) are both easy and exact.
Accuracy degrades the farther x is from a.