tlong nmr 19 dan 20. aq blm faham caranya

jawab

19. 
f(x) = (x² -3x + 5)⁴ (2x - 3)²
f(1) = (1-3+5)⁴(2-3)²
f(1) = 3⁴ (-1)² = 81

f ' (x) = 4(2x-3)(x² -3x +5)³(2x -3)² + 2(2)(2x -3) (x²-3x +5)⁴ 
f' (1) = 4(2-3) (1 - 3 + 5)³ (2-3)² + 2(2)(-1)(1-3+5)⁴
f'(1) = 4(-1)(27)(1) - 4(81)
f' (1) = - 108 - 324 = - 432

f(1) + f'(1) = 81 - 432 = - 351


20. ukuran kotak
p= 30 - 2x
l = 30 - 2x
t = x

V(x) = (30 - 2x)(30-2x)(x)
V(x) = 900x - 120 x² + 4x³
maksimum jika V'(x) = 0
900 - 240x + 12 x² = 0
x² - 20x + 75 = 0
(x - 15)(x - 5) = 0
x = 15 (tolak)  atau x = 5
p= 30- 2x = 30-10 = 20
l = 30-2x = 30-10 = 20
t = 5

V = plt = 20 . 20. 5 = 2.000

19. -351
20. 2.000 cm

maaf klo salah