Berikut ini adalah pertanyaan dari cchocolatteaa pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas
d) Find the coordinates of the point on the curve where the gradient is minimum and state the value of the minimum gradient.
*If possible pls send it together with the working way, thank you so much
Jawaban dan Penjelasan
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Jawab:
y = 2x³+ax²+bx+7 ⇒ substitute with (2, -13)
-13 = 2(2)³+a(2)²+b(2)+7 (I assume it's supposed ax², not ax)
-13 = 16+4a+2b+7
4a+2b=-36
b = -18-2a ⇒ [1]
a.
stationary point ⇒ gradient = 0 (y' = 0)
y = 2x³+ax²+bx+7
y' = 6x² + 2ax +b
0 = 6x² + 2ax +b ⇒ x = 2
0 = 6(2)² + 2a(2) +b ⇒ substitute with [1]
0 = 24 + 4a + (-18-2a)
2a = -6
a = -3
b = -18-2a = -18-2(-3) = -12
b.
y' = 6x² + 2ax +b
0 = 6x² + 2(-3)x +(-12)
0 = 6x² - 6x -12
0 = x² - x - 2
(x-2)(x+1) = 0
x₁ = 2
x₂ = -1
y = 2x³-3x²-12x+7
y(-1) = 2(-1)³-3(-1)²-12(-1)+7
= 14
coordinate (-1, 14)
c. (-1, 14) maximum point
(2, -13) minimum point
d.
y' = gradient
gradient min = y' min
= 6x² - 6x -12 min ⇒ y'' = 0
y'' = 12x - 6
0 = 12x - 6
x = 1/2
y(1/2) = 2(1/2)³-3(1/2)²-12(1/2)+7
= 1/2
coordinate ( ,
)
![Jawab:y = 2x³+ax²+bx+7 ⇒ substitute with (2, -13)-13 = 2(2)³+a(2)²+b(2)+7 (I assume it's supposed ax², not ax)-13 = 16+4a+2b+74a+2b=-36b = -18-2a ⇒ [1]a.stationary point ⇒ gradient = 0 (y' = 0)y = 2x³+ax²+bx+7y' = 6x² + 2ax +b0 = 6x² + 2ax +b ⇒ x = 20 = 6(2)² + 2a(2) +b ⇒ substitute with [1]0 = 24 + 4a + (-18-2a)2a = -6a = -3b = -18-2a = -18-2(-3) = -12b.y' = 6x² + 2ax +b0 = 6x² + 2(-3)x +(-12)0 = 6x² - 6x -120 = x² - x - 2(x-2)(x+1) = 0x₁ = 2x₂ = -1y = 2x³-3x²-12x+7y(-1) = 2(-1)³-3(-1)²-12(-1)+7 = 14coordinate (-1, 14)c. (-1, 14) maximum point (2, -13) minimum pointd. y' = gradientgradient min = y' min = 6x² - 6x -12 min ⇒ y'' = 0y'' = 12x - 60 = 12x - 6x = 1/2y(1/2) = 2(1/2)³-3(1/2)²-12(1/2)+7 = 1/2coordinate ([tex]\frac{1}{2}[/tex] , [tex]\frac{1}{2}[/tex])](https://id-static.z-dn.net/files/dfc/498f8a0df5690c0c2fef60eb83b9b541.png)
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Last Update: Mon, 09 May 22