tolong dibntu. ini soal kisi2 mate untuk semester bsk

17). f(x) = (x + 1)(2x - 3)(3x + 2)
f(x) = (2x² - x - 3)(3x + 2)
f(x) = 6x³ + x² - 11x - 6
f'(x) = 3.6x³⁻¹ + 2.x²⁻¹ - 11x¹⁻¹ - 0
f'(x) = 18² + 2x - 11 (B)

18). f(x) = x⁴ - 4x²
Nilai minimum diperoleh ketika f'(x) = 0
f'(x) = 4x³ - 8x ⇔ 4x(x² - 8) = 0
⇔ 4x(x - √8)(x + √8) = 0
x = 0 atau x = √8 dan x = -√8
f(x) = x⁴ - 4x²
f(0) = 0
f(-√8) = 32
f(√8) = 32
Jadi, nilai minimum = 0 (E)

19). f(x) = x³ - 9x² + 15x, interval -1 < x < 6
1). Titik balik diperoleh ketika f'(x) = 0
f'(x) = 3x² - 18x + 15 = 0
f'(x) = x² - 6x + 5 = 0
⇒ (x - 5)(x - 1) = 0
⇒ x = 5 atau x = 1

Nilai balik:
f(x) = x³ - 9x² + 15x
f(5) = 5³ - 9(5)² + 15(5) = - 25
f(1) = 1³ - 9(1)² + 15(1) = 7

Nilai pada ujung interval:
f(x) = x³ - 9x² + 15x
f(-1) = -1³ - 9(-1)² + 15(-1) = -25
f(6) = 6³ - 9(6)² + 15(6) = -18

Jadi, nilai maksimumnya adalah -25 (B)

17) f(x) = (x + 1)(2x - 3)(3x + 2)
= (x + 1)(6x^2 + 4x - 9x - 6)
= (x + 1)(6x^2 - 5x - 6)
= 6x^3 - 5x^2 - 6x + 6x^2 - 5x - 6
= 6x^3 + x^2 - 11x - 6
f'(x) = 18x^2 + 2x - 11

18) f(x) = x^4 - 4x^2
f'(x) = 4x^3 - 8x = 0
=> 4x(x^2 - 2) = 0
=> 4x(x + √2)(x - √2) = 0
=> x = 0, x = -√2, x = √2
x = 0 => f(0) = 0^4 - 4(0)^2 = 0
x = √2 => f(√2) = (√2)^4 - 4(√2)^2 = 4 - 4(2) = -4
x = -√2 => f(-√2) = (-√2)^4 - 4(-√2)^2 = 4 - 4(2) = -4
Jadi nilai balik minimumnya = -4 (tak ada di option)

3) f(x) = x^3 - 9x^2 + 15x pd interval -1 < x < 6
f'(x) = 3x^2 - 18x + 15 = 0
=> x^2 - 6x + 5 = 0
=> (x - 1)(x - 5) = 0
=> x = 1 atau x = 5
x = 1 => f(1) = 1^3 - 9(1)^2 + 15(1) = 1 - 9 + 15 = 7
x = 5 => f(5) = 5^3 - 9(5)^2 + 15(5) = 125 - 225 + 75 = -25
Jadi nilai MAKSIMUM nya = 7 (tak ada di option)