easy rider
New member
The question is to determine the gradient of the curve, y=3x(2x-3)², at point (2, 6).
The answer in the book is 27, and the answer from a graphic calculator is 27. I got -45 though, here's my working:
y=3x(2x-3)²
y=3x(2x²-12x+9)
f '(x)= 3x(4x-12) + 3(2x²-12x+9)
. . .= 12x² - 36x + 6x² - 36x + 27
. . . = 18x² - 72x + 27
therefore gradient = 18(2)² - 72(2) + 27
. . . . . . . . . . . . . = -45
aah thanks it was something obvious
The answer in the book is 27, and the answer from a graphic calculator is 27. I got -45 though, here's my working:
y=3x(2x-3)²
y=3x(2x²-12x+9)
f '(x)= 3x(4x-12) + 3(2x²-12x+9)
. . .= 12x² - 36x + 6x² - 36x + 27
. . . = 18x² - 72x + 27
therefore gradient = 18(2)² - 72(2) + 27
. . . . . . . . . . . . . = -45
aah thanks it was something obvious