17. Differentiation (IV) - task solutions

MATH1110 Lecture tasks - Worked solutions Differentiation (IV) Task 1. 1. Using the data given in the question, we have f 0 (0) = 0 - 0 + 2 0 - 0 + 1 = 2 . Since f 0 (0) > 0, the function f is increasing , and therefore the statement is FALSE. 2. From this point on it becomes much simpler if we instead locate the stationary points of the function and use these to determine the function's nature about each. Solving f 0 ( x ) = 0 gives 2 x 2 - 6 x + 2 x 4 - 2 x 2 + 1 = 0 = ⇒ 2 x 2 - 6 x + 2 = 0 = ⇒ x = 6 ± √ 20 4 , and so we have stationary points at x = 1 2 (3 + √ 5) and x = 1 2 (3 - √ 5). Note that the function (and therefore its derivative) is also undefined at x = ± 1. Using the first derivative test, we find the following: x - 2 - 1 0 1 2 (3 - √ 5) 1 2 1 2 1 2 (3 + √ 5) 3 f 0 ( x ) 22 9 DNE 2 0 - 8 9 DNE - 2 9 0 1 32 Slope / / — \ \ — / Nature inc. inc. max. dec. dec. min. inc. We find that f is increasing for - 5 < x < - 1, and therefore the statement is TRUE. 3. f has a stationary point at x = 1 2 (3 + √ 5), and therefore the statement is TRUE. Task 2. The stationary points in the interval [0 , 2 π ] occur when p 0 ( x ) = 0 = ⇒ 1 - 2 cos( x ) = 0 = ⇒ cos( x ) = 1 2 = ⇒ x = π 3 , 5 π 3 . Using the first derivative test, we find the following: x 0 π 3 π - 5 π 3 2 π f 0 ( x ) - 1 0 3 0 - 1 Slope \ — / — \ Nature dec. min. inc. max. dec. That is, p ( x ) has a local minimum at x = π/ 3 and a local maximum at x = 5 π/ 3. Page 1

MATH1110 Lecture tasks - Worked solutions Differentiation (IV) Task 3. Let us define g ( u ) = √ u and u ( t ) = 1 + 4 t 2 so that g ( u ( t )) = √ 1 + 4 t 2 . Then g 0 ( u ) = 1 2 u - 1 / 2 and u 0 ( t ) = 8 t . Thus the Chain Rule gives us g 0 ( t ) = dg du du dt = 1 2 √ u 8 t = 4 t √ 1 + 4 t 2 . We need to differentiate this one more time, but this time we will require the Quotient Rule. Let us define u ( t ) = 4 t and v ( t ) = √ 1 + 4 t 2 so that u ( t ) v ( t ) = 4 t √ 1 + 4 t 2 . Then u 0 ( t ) = 4 and v 0 ( t ) = 4 t √ 1 + 4 t 2 (using the above), so that g 00 ( t ) = d dt ( g 0 ( t ) ) = d dt u ( t ) v ( t ) = v ( t ) u 0 ( t ) - u ( t ) v 0 ( t ) ( v ( t ) ) 2 = √ 1 + 4 t 2 · 4 - 4 t · 4 t √ 1+4 t 2 ( √ 1 + 4 t 2 ) 2 = 4 ( √ 1 + 4 t 2 ) 2 - 16 t 2 ( √ 1 + 4 t 2 ) 3 = 4 ( √ 1 + 4 t 2 ) 3 . Task 4. 1. Using the data given in the question, we have f 00 (2) = 4(2) 3 - 18(2) 2 + 12(2) - 6 (1 - 2 2 ) 3 = 22 27 . Since f 00 (2) > 0, the function f is concave up, and therefore the statement is TRUE. 2. From this point on it becomes much simpler if we instead locate the points of inflection of the function and use these to determine the function's concavity about each. Solving f 00 ( x ) = 0 (which is beyond the scope of this course) gives 4 x 3 - 18 x 2 + 12 x - 6 (1 - x 2 ) 3 = 0 = ⇒ 4 x 3 - 18 x 2 + 12 x - 6 = 0 = ⇒ x = ( 3 √ 5 ) 2 + 3 √ 5 + 3 2 , and so we have a single point of inflection at x ≈ 3 . 817. Note that the function (and therefore its second derivative) is also undefined at x = ± 1. Examining the second Page 2

MATH1110 Lecture tasks - Worked solutions Differentiation (IV) derivative at various points, we find the following: x - 2 - 1 0 1 2 3 . 817 4 f 00 ( x ) 134 27 DNE - 6 DNE 22 27 0 - 2 675 Concavity ∪ ∩ ∪ — ∩ Nature conc. up conc. down conc. up POI conc. down We find f is concave down for - 1 < x < 0, and therefore the statement is TRUE. Task 5. Using the Chain Rule, we have f 0 ( x ) = e 1 - 2 x 2 · ( - 4 x ) = - 4 xe 1 - 2 x 2 . To compute the second derivative, we will need to use the Product Rule to obtain f 00 ( x ) = e 1 - 2 x 2 · ( - 4) + ( - 4 x ) · - 4 xe 1 - 2 x 2 = (16 x 2 - 4) e 1 - 2 x 2 . The stationary points of f occur when f 0 ( x ) = 0. That is, when - 4 xe 1 - 2 x 2 = 0 = ⇒ x = 0 . We find f 00 (0) = - 4 e < 0, and so by the second derivative test, f has a local maximum stationary point at (0 , f (0)) = (0 , e ). Task 6. f is the black curve (B), f 0 is the blue curve (A), and f 00 is the red curve (C). To see this, note that the stationary points of f occur at the same x -values as the x -intercepts of f 0 , and the stationary points of f 0 occur at the same x -values as the x -intercepts of f 00 . It's also possible to make additional observations about the various curves being increasing/decreasing on certain intervals, and similarly regarding their concavity, but these are not necessary to uniquely identify which curve corresponds to which function. Task 7. Using the Product Rule (and also the Chain Rule), we obtain h 0 ( t ) = √ t + 3 · 1 + t · 1 2 √ t + 3 = 3 t + 6 2 √ t + 3 . The stationary points of h will occur when h 0 ( t ) = 0. That is, when 3 t + 6 2 √ t + 3 = 0 = ⇒ 3 t + 6 = 0 = ⇒ t = - 2 . Using the first derivative test, we obtain x - 2 . 1 - 2 - 1 . 9 h 0 ( x ) 3( - 2 . 1) + 6 2 √ - 2 . 1 + 3 < 0 0 3( - 1 . 9) + 6 2 √ - 1 . 9 + 3 > 0 Slope \ — / Nature decreasing min. increasing Page 3