Find the point at which the tangent to the curve y = √(4x - 3) - 1

1.	Find the point at which the tangent to the curve y = √(4x - 3) - 1 has its slope 2/3.
Answer» Given curve is y = √(4x - 3) - 1 Differentiating, we have dy/dx = 1/2√(4x - 3) x 4 = 2/√(4x - 3) Also, slope of tangent at any point on the curve is 2/3 ∴ dy/dx = 2/3 2/√(4x - 3) = 2/3 √(4x - 3) = 3 4x - 3 = 9 4x = 12 ∴ x = 3 So, y = √(4 x 3 - 3) - 1 = √9 - 1 = 2 So, the require point is (3, 2)

Find the point at which the tangent to the curve y = √(4x - 3) - 1 has its slope 2/3.

Answer»

Given curve is y = √(4x - 3) - 1

Differentiating, we have

dy/dx = 1/2√(4x - 3) x 4 = 2/√(4x - 3)

Also, slope of tangent at any point on the curve is 2/3

∴ dy/dx = 2/3

2/√(4x - 3) = 2/3

√(4x - 3) = 3

4x - 3 = 9

4x = 12

∴ x = 3

So, y = √(4 x 3 - 3) - 1 = √9 - 1 = 2

So, the require point is (3, 2)