180 Chapter 5. Graphs and the Derivative (LECTURE NOTES 11) 5.3 Higher Derivatives, Concavity, and the Second Derivative Test Notation for higher derivatives of y = f(x) include ′′ d2y 2 second derivative: f (x),, dx2 ,Dx[f(x)], ′′′ d3y 3 third derivative: f (x),, dx3 ,Dx[f(x)], n (n) d( )y (n) n f x ,, n ,D f x . fourth and above, th, derivative: ( ) dx( ) x [ ( )] Second derivative of function can be used to check for both concavity and points of inflection of the graph of function. y concave up f’’(x) > 0 point of maximum in!ection x concave down minimum f’’(x) < 0 Figure 5.9 (Concave up, concave down and point of inflection) In particular, if function f has both derivatives f ′ and f ′′ for all x in (a, b), the f(x) is concave up if f ′′(x) > 0, for all x in (a,b) f(x) is concave down if f ′′(x) < 0, for all x in (a,b) At an inflection point of function f, either f ′′(x) = 0 or second derivative does not exist (although the reverse is not necessarily true). Second derivative test is used to check for relative extrema. Let f ′′ exist on open interval containing c (except maybe c itself) and let f ′(c) = 0, then if f ′′(c) > 0 then f(c) is relative minimum if f ′′(c) < 0 then f(c) is relative maximum if f ′′(c)=0or f ′′(x) does not exist test gives no information, use first derivative test Exercise 5.3 (Higher Derivatives, Concavity, and the Second Derivative Test) Section 3. Higher Derivatives, Concavity, and the Second Derivative Test (LECTURE NOTES 11)181 1. Concave up, concave down and inflection points. y 10 vertical tangent line at point C 8 D F 6 C A I G 4 H B 2 E 0 -5 -3 -1 1 3 5 x Figure 5.10 (Concave up, concave down and inflection points) (a) Function is concave up between points (choose one or more) (i) A to C (ii) C to E (iii) E to F (iv) F to G (v) G to I (vi) none because slope always increases from left endpoint to right endpoint, without “holes” (b) Function is concave down between points (choose one or more) (i) A to C (ii) C to E (iii) E to F (iv) F to G (v) G to I (vi) none because slope always decreases from left endpoint to right endpoint (c) Inflection point(s) at (choose one or more) (i) A (ii) B (iii) C (iv) D (v) E (vi) F (vii) G (viii) H (ix) I (x) none because, at point C, concavity flips from up to down, although f ′′ does not exist and, at point G, concavity flips from down to up and f ′′ = 0 2. More concave up, concave down and inflection points. 182 Chapter 5. Graphs and the Derivative (LECTURE NOTES 11) tangent line NOT vertical y C 10 at point D 8 G D B F 6 A vertical tangent line 4 at endpoint H E H 2 0 -5 -3 -1 1 3 5 x Figure 5.11 (Concave up, concave down and inflection points) (a) Function is concave up between points (choose one or more) (i) A to B (ii) B to C (iii) C to F (iv) D to F (v) F to H (vi) none because slope always increases from left to right endpoints (b) Function is concave down between points (choose one or more) (i) A to B (ii) B to C (iii) C to F (iv) D to F (v) F to H (vi) none because slope always decreases from left endpoint to right endpoint (c) One inflection point at (i) A (ii) B (iii) C (iv) D (v) E (vi) F (vii) G (viii) H because, at point F, concavity flips from up to down and f ′′ = 0, but, at point D, concavity remains down, so f ′′ =6 0, and, at point B, function does not exist, so cannot be an inflection point 3. Examples of higher derivatives. (a) f(x)= x2 +4x 21 − So f ′(x)=2x2−1 + 4(1)x1−1 = (i) 2x2 +4 (ii) 2x − 21 (iii) 2x +4 Section 3. Higher Derivatives, Concavity, and the Second Derivative Test (LECTURE NOTES 11)183 and f ′′(x) = 2(1)x1−1 = (i) 2x (ii) 2 (iii) 4 and so f ′′(3) = (i) 3 (ii) 6 (iii) 2 and so f ′′( 8) = (i) 2 (ii) 3− (iii) 6 (b) f(x)=2x3 +3x2 36x − So f ′(x) = 2(3)x3−1 + 3(2)x2−1 36(1)x1−1 = 3 2 − 2 (i) 6x +6 (ii) 6x +6x − 36x (iii) 6x +6x − 36 and f ′′(x) = 6(2)x2−1 + 6(1)x1−1 = (i) 12x +6 (ii) 12x2 +6x (iii) 12+6x and so f ′′(3) = 12(3)+6 = (i) 36 (ii) 32 (iii) 42 and so f ′′( 8) = 12( 8)+6= (i) −45 (ii)− −90 −(iii) −8 and f ′′′(x)= (i) 12 (ii) 6x (iii) 12x and so f ′′′(3) = (i) 12 (ii) 3 (iii) 4 and f (4)(x)= (i) 0 (ii) 12 (iii) 12x (c) f(x)=3x2 2x4 − So f ′(x) = 3(2)x2−1 2(4)x4−3 = 3 − 2 3 2 (i) 6x − 8x (ii) 6x +2x (iii) 6x +6x d2y 1−1 3−1 and dx2 = 6(1)x 8(3)x = 2− (i) 6 (ii) 6 − 24x (iii) 6x 3 2−1 and Dx[f(x)] = 24(2)x (i) 6x (ii) −12− (iii) −48x 184 Chapter 5. Graphs and the Derivative (LECTURE NOTES 11) and f (4)(x)= (i) 12 (ii) 48x (iii) −48 (d) f(x)= √3x 3 − Let f[g(x)] = (3x 3)1/2, where g(x)=3x 3, and f(x)= x1/2 − 1 − ′ ′ 1 2 −1 1−1 so Dx[f(x)] = f [g(x)]g (x)= 2 (3x 3) 3(1)x = 3 − 1 3 1− 3 × − 3 2 2 2 (i) 2 (3x − 3) (ii) 2 (3x − 3) (iii) 2 (3x − 3) 1 1 3 − 2 3 − 2 Let f[g(x)] = 2 (3x 3) , where g(x)=3x 3, and f(x)= 2 x − − 1 2 ′ ′ 3 1 − 2 −1 so Dx[f(x)] = f [g(x)]g (x)= 2 ( 2 (3x 3) 3= 9 − 3 3 −3 − 9 × − 3 2 2 2 (i) 4 (3x − 3) (ii) − 4 (3x − 3) (iii) − 4 (3x − 3) 3 3 9 − 2 9 − 2 Let f[g(x)] = 4 (3x 3) , where g(x)=3x 3, and f(x)= 4 x − − − 3 − 3 ′ ′ 9 3 − 2 −1 so Dx[f(x)] = f [g(x)]g (x)= 4 ( 2 (3x 3) 3= 27 − 3 81 − −5− −81 × − 5 2 2 2 (i) 8 (3x − 3) (ii) 8 (3x − 3) (iii) − 8 (3x − 3) (e) f(x)=5e2x Let f[g(x)]=5e2x, where g(x)=2x, and f(x)=5ex ′ ′ 2x 1−1 so Dx[f(x)] = f [g(x)]g (x)=5e 2(1)x = 2x 2 2x × (i) 5e (ii) 5e (iii) 10e Let f[g(x)] = 10e2x, where g(x)=2x, and f(x)=10ex 2 ′ ′ 2x 1−1 so Dx[f(x)] = f [g(x)]g (x)=10e 2(1)x = (i) 20e2x (ii) 10e2x (iii) 20ex × Let f[g(x)] = 20e2x, where g(x)=2x, and f(x)=20ex 3 ′ ′ 2x 1−1 so Dx[f(x)] = f [g(x)]g (x)=20e 2(1)x = 2x 2x 2x × (i) 10e (ii) 40e (iii) 5e (f) f(x) = tan2x. Recall, derivatives of trigonometric functions include: D [sin x] = cos x D [csc x]= cot x csc x x x − D [cos x]= sin x D [sec x] = tan x sec x x − x D [tan x] = sec2 x D [cot x]= csc2 x x x − So, let f[g(x)] = tan2x, where g(x)=2x, and f(x) = tan x so D [f(x)] = f ′[g(x)]g′(x) = sec2 2x 2(1)x1−1 = x × Section 3. Higher Derivatives, Concavity, and the Second Derivative Test (LECTURE NOTES 11)185 (i) 2 sec 2x (ii) 2 sec2 2x (iii) 2 sec x Let f[g(x)] = 2 sec2 2x, where g(x) = sec 2x, and f(x)=2x2 2 ′ ′ 2−1 so Dx[f(x)] = f [g(x)]g (x) = 2(2) sec 2x 2tan2x sec 2x = (i) 8tan2x sec2 2x (ii) 6tan2x sec2 2x ×(iii) 8tan2x sec 2x 4. Application: throwing a ball. A ball is thrown upwards with an initial velocity of 32 feet per second and from an initial height of 150 feet. A function relating height, f(t), to time, f, when throwing this ball is: f(t)= 12t2 + 32t + 150 − Find velocity, v(t)= f ′(t), acceleration, a(t)= v′(t)= f ′′(t), at t = 5 seconds. So f ′(t)= 12(2)t2−1 + 32(1)t1−1 = − (i) 24t + 32 (ii) −24t (iii) −24t + 32 and so f ′(5) = 24(5)+32 = (i) 88 (ii) 5 −(iii) −88 feet per second and f ′′(t)= 24(1)t1−1 = (i) −24 (ii)−24 (iii) −88 and so f ′′(5) = (i) 24 (ii) −24 (iii) −88 feet per second2 True / False. Acceleration is first derivative of velocity or second derivative of height. 5. Application: index of absolute risk aversion. Index of absolute risk aversion is U ′′(M) I(M)= − , U ′(M) where M is quantity of a commodity owned by a consumer and U(M) is utility (fulfillment) a consumer derives from the quantity M of the commodity. Find I(M) if U(M)= M 3.