# MA1M01 Calculus Assignment 4

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MA1M01 Calculus Assignment 4

MA1M01 Calculus Assignment 4 1. Differentiate the following: (a) f (x) = sin2 (x2 ) d d d f (x) = 2 sin(x2 ) · sin(x2 ) = 2 sin(x2 ) cos(x2 ) x2 dx dx dx = 2 sin(x2 ) cos(x2 )2x = 4x sin(x2 ) cos(x2 ) (b) g(x) = cos(x3 ) sin(x2 + 3x) d d d g(x) = cos(x3 ) · sin(x2 + 3x) + cos(x3 ) · sin(x2 + 3x) dx dx dx = cos(x3 ) cos(x2 + 3x) · (2x + 3) − sin(x3 ) · 3x2 · sin(x2 + 3x) = (2x + 3) cos(x3 ) cos(x2 + 3x) − 3x2 sin(x3 ) sin(x2 + 3x) (c) h(x) = sin (Note that h(x) = sin g(x) f (x) cos(x3 ) sin(x2 + 3x) sin2 (x2 ) ) 1 g(x) d g(x) · f (x) dx f (x) g(x) f (x)g 0 (x) − f 0 (x)g(x) = cos · f (x) [f (x)]2 cos(x3 ) sin(x2 + 3x) · = cos sin2 (x2 ) (sin2 (x2 ))(2x + 3)[cos(x3 ) cos(x2 + 3x) − 3x2 sin(x3 ) sin(x2 + 3x)] sin4 (x2 ) 4x sin(x2 ) cos(x2 )(cos(x3 ) sin(x2 + 3x)) − sin4 (x2 ) d h(x) = cos dx 2. Consider the following function: f (x) = 25x2 (x − 1)(x − 2)2 for − 0.1 < x < 2.1 (a) Find all critical points of f (x). (Hint: Differentiate using the produst rule and factorise.) The critical points occur when f 0 (x) = 0 (or when f 0 (x) doesn’t exist, but that doesn’t happen here). d f (x) = dx = = = 50x(x − 1)(x − 2)2 + 25x2 (x − 2)2 + 50x2 (x − 1)(x − 2) 25x(x − 2)[2(x − 1)(x − 2) + x(x − 2) + 2x(x − 1)] 25x(x − 2)[2x2 − 6x + 4 + x2 − 2x + 2x2 − 2x] 25x(x − 2)[5x2 − 10x + 4] Solve the qudratic: x = 10 ± √ 10 ± 2 5 1 100 − 80 = =1± √ 10 10 5 √ So we have 4 critical points: 1 1 x = 0, 2, 1 − √ , 1 + √ 5 5 (b) Which critical points are relative minima? Which are relative maxima? By the previous part we have: 1 1 f 0 (x) = 25x(x − 2)(x − (1 − √ ))(x − (1 − √ )) 5 5 (i) for x < 0, all 4 factors are negative so f 0 (x) > 0 2 (ii) for 0 < x < 1 − (iii) for 1 − (iv) for 1 + √1 5 √1 5 √1 , 5 three of the factors are negative so f 0 (x) < 0 <x<1+ √1 , 5 two of the factors are negative so f 0 (x) > 0 < x < 2, one of the factors are negative so f 0 (x) < 0 (v) for x > 2 all of the facrors are positive so f 0 (x) > 0 You could also determine the sign of f 0 (x) by picking any point on the above intervals and looking at the sign of f ’ on this point. Using the first derivative test: • (i)+(ii) and (iii)+(iv) tells us that x = 0 and x = 1 + √15 are relative maxima • (ii)+(iii) and (iv)+(v) tells us that x = 1 − √15 and x = 2 are relative minima (Alternatively we could have used the second derivative test.) (c) Which of the relative minima is smallest? Which of the relative maxima is largest? 1 16 1 16 f (0) = 0, f (1 − √ ) = − √ ≈ −7.155, f (1 + √ ) = √ ≈ 7.155, f (2) = 0 5 5 5 5 So the largest relative maximum of f (x) is at x = 1 + √15 with a value smallest relative minimum is at x = 1 − √15 with a value of − √165 . 3 16 √ and 5 the