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2011-05-09 6:37 pm

Please show all workings: (Quick)Q1) Find the equation of the tangent to the curve y=x-In x at the point where x=2. At which point does the tangent cut the x-axis?Q2) Find the equation of the tangent to the curve y=sin x at the point where x = ╥/4. At which point does this tangent cut the x-axis? * ╥ =3.14

回答 (1)

用戶25975

2011-05-09 6:47 pm

✔ 最佳答案

1) dy/dx = 1 - 1/x

So at x = 2, dy/dx = 1 - 1/2 = 1/2

By point-slope form, the eqn of tangent is:

[y - (2 - ln 2)]/(x - 2) = 1/2

2y - 2(2 - ln 2) = x - 2

x - 2y + (2 - 2 ln 2) = 0

With y = 0, x = 2 ln 2 - 2 and hence the tangent cuts the x-axis at (2 ln 2 - 2, 0)

2) dy/dx = cos x

At x = π/4, dy/dx = √2/2

By point-slope form, the eqn of tangent is:

(y - √2/2)/(x - π/4) = √2/2

2y - √2 = x√2 - (√2π)/4

x√2 - 2y + [√2 - (√2π)/4] = 0

With y = 0, x = π/4 - 1 and hence the tangent cuts the x-axis at (π/4 - 1, 0)

參考: 原創答案

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