What is the slope of the line normal to the tangent line of f(x) = 2x-4sqrt(x-1)  at  x= 2 ?

Feb 22, 2017

The slope will be undefined.

Explanation:

Start by finding the y-coordinate of the point of tangency.

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

$f \left(2\right) = 4 - 4$

$f \left(2\right) = 0$

Find the derivative of $f \left(x\right)$.

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

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

Now find the slope of the tangent.

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

The normal line is perpendicular to the tangent line. The slope of $0$ of the tangent line means the line will be $y = a$, where $a$ is a constant. Then the line perpendicular to this will be of the form $x = b$, where the slope is undefined.

Hopefully this helps!