# What is the slope of the line normal to the tangent line of f(x) = 2x-xsqrt(x-1)  at  x= 2 ?

Mar 6, 2018

The slope of the normal line to the tangent line is undefined.

#### Explanation:

$f \left(x\right) = 2 x - x \sqrt{x - 1}$

$f ' \left(x\right) = 2 - \sqrt{x - 1} - \frac{x}{2 \sqrt{x - 1}}$

$f ' \left(2\right) = 2 - \sqrt{2 - 1} - \frac{2}{2 \sqrt{2 - 1}} = 0$

So the slope of the tangent line is 0. This is equivalent to saying that the tangent line is horizontal.

$f \left(2\right) = 2 \left(2\right) - \left(2\right) \sqrt{2 - 1} = 4 - 2 = 2$

Since the y-coordinate of the function is 2, the horizontal tangent line must be defined by the equation y = 2.

Since they want the slope of the normal line to the tangent line, the normal line would have to be a vertical line since the tangent line is a horizontal line.

Hence, the normal line to the tangent must be a vertical line with an equation of x = 2. This means that the slope of the normal line to the tangent line is undefined. Vertical lines have no change in x, which results in an undefined form (since denominator of slope computation is 0).