✔ 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)