4.3 – Derivatives and the shapes of curves The Mean Value Theorem: If f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that: b a c

Increasing / Decreasing Test If f’(x) > 0 on an interval, then f is increasing on that interval. If f’(x) < 0 on an interval, then f is decreasing on that interval. ex: Find where the function f(x) = 3x4 – 4x3 – 12x2 + 5 is increasing and decreasing… f ’(x) = 12x3 – 12x2 – 24x Factor: 12x(x – 2)(x + 1) Critical Numbers: x = -1, 0, 2 Test: ( - , -1) (-1, 0) (0, 2) (2, )

The First Derivative Test For c being a critical number of a continuous function f… If f ’ changes from positive to negative at c, then f has a local maximum at c. If f ’ changes from negative to positive at c, then f has a local minimum at c. If f ’ does not change sign at c, then f has no local maximum or minimum at c. c2 c1 f ’ FROM – TO + = MIN FROM + TO – = MAX

ex: Critical Numbers: x = -1, 0, 2 Test: ( - , -1) (-1, 0) (0, 2) (2, ) MIN @ -1 MAX @ 0 MIN @ 2 Concavity Test If f ’’(x) > 0 in [a,b], then f is concave up on [a,b] If f ’’(x) < 0 in [a,b], then f is concave down on [a,b]

The Second Derivative Test For f ’’ being continuous near c… If f ’(c) = 0 and f ’’(c) > 0 then f has a local minimum at c. If f ’(c) = 0 and f ’’(c) < 0 then f has a local maximum at c. f ’ f ’’ c1 c2 f

ex: For the curve f(x) = x4 - 4x3 use calculus to find local maxima and minima, concavity intervals and points of inflection… Critical #’s = 0, 3 [ Test f ’ ] (- , 0) f’(-1) = No Max / Min @ 0 f’(1) = (0, 3) (3, ) f’(4) = + NegPos = Min @3 Local Min

+ + Test second derivative for concavity intervals: [ Test f ’’ ] 2nd derivative test inconclusive @ f ’ critical # 0 [ Test f ’’ ] (- , 0) f ’’(-1) = + f ’ critical # 3 = (36)(1) = 36, so f ’’ is + = MIN @ 3 CC Inflection point @ 0 f ’’(1) = (0, 2) CC Inflection point @ 2 (2, ) f ’’(5) = + CC Summary Local min at (3, -27) No local max CC up from (– , 0) U (2, ) CC down from (0, 2)

f(x) = x4 - 4x3

nDeriv For f(x) = x2/3(6 – x)1/3 graph the function and its first and second derivatives…. Type f(x) into Y1 Go to Y2 line Go to MATH [8] (nDeriv) / Press ENTER Set nDeriv arguments to (Y1, X, X) *Y1 in VARS ~ Y-VARS ~ FUNCTION ~ #2 Repeat on Y3 line using (Y2, X, X) for nDeriv Graph