Jawaban Terbaik apabila memfotokan cara kerja, Mohon bantuannya kakak :)

jawab

g(x) = f(2x-1)
g(x) = 1/3 (2x-1)³ - A²(2x-1) - 7
g'(x) = 2 (2x -1)² - 2A² 

syarat maksimum --> g'(x) = 0 dan g" (x)  < 0
2(2x-1)² - 2A² = 0
2 { (2x -1)² - A²} = 0
(2x - 1)² - A² = 0
(2x - 1 + A)(2x -1- A) = 0
2x - 1 + A = 0
x = 1/2 (1 - A) ...(i)

2x-1 - A= 0
2x = 1 + A
x= 1/2 (1 + A)...(ii)

g'(x) =2(2x-1)² - A²
g"(x) = 4(2)(2x-1)
g"(x)= 8x - 8
pada interval -1/2 ≤ x ≤ 3/2
x = - 1/2
g"(-1/2) = 8(-1/2) - 8
g"(-1/2) = - 12 
g"(-1/2) < 0 titik maksimum

untk x = -1/2
(i)
x= 1/2 (1 - A) --> -1/2 = 1/2 (1  - A)
1  - A = - 1
A= 2
g(x) = 1/3 (2x - 1)³ - A²(2x- 1) - 7
g(x) = 1/3 (2x -1)³ - 4(2x-1)- 7
g(-1/2)= 1/3(-2)³ - 4(-2) - 7 = -5/3