Question

find the equation of the tangent line to the curve at the given point.
$y = 3x - 2\sqrt{x}$, $(1, 1)$
$y = \square$

Explanation:

Step1: Compute derivative of y

Rewrite $y=3x-2x^{1/2}$. Derivative: $y'=3 - 2*(1/2)x^{-1/2}=3 - \frac{1}{\sqrt{x}}$

Step2: Find slope at x=1

Substitute x=1 into $y'$: $y'(1)=3 - \frac{1}{\sqrt{1}}=2$

Step3: Use point-slope form

Point-slope: $y - y_1 = m(x - x_1)$. Substitute (1,1) and m=2: $y -1=2(x -1)$

Step4: Simplify to slope-intercept

Expand: $y -1=2x -2$ → $y=2x -1$

Answer:

$2x - 1$