The point at which the tangent to the curve \(y=\sqrt{4x-3}-1\) has its slope \(\displaystyle\frac{2}{3}\), is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (3, 2)
The slope of the tangent is dy/dx. By the chain rule:
\( \displaystyle \frac{dy}{dx} = \frac{1}{2\sqrt{4x-3}}\times 4 = \frac{2}{\sqrt{4x-3}} \)
Setting it equal to \(\frac{2}{3}\):
\(\sqrt{4x-3} = 3\)
\(4x - 3 = 9\)
\(x = 3\)
Then
\(y = \sqrt{9} - 1 = 2\)
So the point is (3, 2).