AB Calculus - Unit 6

(4/3) × π × r³

4 × π * r²

(1/3) × π × r² × h

π × r² × h

if lim[x → a] [f(x) / g(x)] = 0 / 0 or ∞/∞, then lim[x → a] [f(x) / g(x)] = lim[x → a] [f’(x) / g’(x)]

find derivative of f(c) and value of f(c); write linear equation

if underestimate, f(x) us concave up; if overestimate, f(x) is concave down

if f’(x) changes from negative to positive at c, f(c) is a relative minimum; if f’(x) changes from positive to negative at c, f(c) is a relative maximum;

if f’’(c) > 0, f(c) is a relative minimum; if f’’(c) < 0, f(c) is a relative maximum

if f’(x) is increasing, f(x) is concave upward; if f’(x) is decreasing, f(x) is concave downward

if f’’(x) > 0, f(x) is concave upward; if f’’(x) < 0, f(x) is concave downward

f’’(c) = 0 or dne; f’’ changes sign (or from increasing to decreasing) at x = c

if f(x) is continuous on [a, b] and differentiable on (a, b) and f(a) = f(b), then there must be a point f’(c) = 0

if f(x) is continuous on [a, b] and differentiable on (a, b), then there must be a point f’(c) = [f(b) - f(a)] / [b - a]

set f’(x) to zero and solve; find where f’(x) is undefined

find f(x) values of critical values; find f(x) values of interval endpoints; largest f(x) value is absolute maximum, smallest f(x) value is absolute minimum

identify variables; find two equations; sub in one equation (only 2 variables); consider domain and use candidates test