Differentiation 1-Methods

1.calculate the gradient of the curve y=x^4/3 at the point where x=8

If you mean y=x4/3

dy/dx = d(x4/3)/dx
dy/dx = 4x³/3 ......... (*)

Substitute x = 8 into (*)
dy/dx = 4(8)³/3
dy/dx = 2048/3

So the gradient of curve at x=8 is 2048/3

If you mean y=x4/3

dy/dx = d(x4/3)/dx
dy/dx = (4/3)x4/3-1
dy/dx = (4/3)x1/3 ......... (*)

Substitute x = 8 into (*)
dy/dx = (4/3)(8)1/3
dy/dx = (4/3)(2)
dy/dx = (8/3)

So the gradient of curve at x=8 is 8/3

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2.use caclus to find the exact coordinatates of the minimum point of the graph of f(x) = 12x²-16x+5, xŒ

f(x) = 12x²-16x+5

f'(x) = 12(2x) - 16
f'(x) = 24x - 16

As coefficient of x² > 0, so the turning point of f(x) is a minimum point.
At minimum point, f'(x) = 0
24x-16 = 0
3x-2 = 0
x = 2/3

when x = 2/3,
f(2/3) = 12(2/3)² - 16(2/3) + 5
f(2/3) = 12(4/9) - 16(2/3) + 5
f(2/3) = 48/9 - 96/9 + 45/9
f(2/3) = -1/3

So the coordinate of the minimum point is (2/3, -1/3).

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3.use calculus to find the values of x for which y=3x4+8x³+6x²-1 has stationary points

y = 3x4 + 8x³ + 6x² - 1

dy/dx = d(3x4 + 8x³ + 6x² - 1)/dx
dy/dx = 3(4x³) + 8(3x²) + 6(2x)
dy/dx = 12x³ + 24x² + 12x
dy/dx = 12x(x² + 2x + 1)
dy/dx = 12x(x+1)² ........ (*)

When y = 3x4 + 8x³ + 6x² - 1 has stationary point,
dy/dx = 0
12x(x+1)² = 0 ......... by (*)
x(x+1)² = 0
x = 0 or x+1 = 0
x = 0 or x = -1

So when x = 0 or x = -1, y = 3x4 + 8x³ + 6x² - 1 has stationary points.

2006-12-10 23:38:23 補充：
小小補充：第一題因為我不知你的意思是 x^(4/3) 或是 (x^4)/3 所以我兩個可能性也計算了，視乎你題目需要選擇適當的答案。