# What is the slope of the line normal to the tangent line of f(x) = x^2-3sqrtx  at  x= 3 ?

May 1, 2018

$\text{Slope of Normal} = - \frac{24 + 2 \cdot \sqrt{3}}{141}$

#### Explanation:

Rewrite the square root: $\sqrt{x} = {x}^{\frac{1}{2}}$

Apply the power rule twice to find the first derivative of $f \left(x\right)$.

$f ' \left(x\right) = 2 x - 3 \cdot \frac{1}{2} \cdot {x}^{- \frac{1}{2}}$
$f ' \left(x\right) = 2 x - \frac{3}{2 \sqrt{x}}$

Evaluating $f ' \left(x\right)$ at $x = 3$ gives the slope of the tangent

$f ' \left(3\right) = 2 \cdot 3 - \frac{3}{2 \sqrt{3}} = \frac{1}{2} \cdot \left(12 - \sqrt{3}\right)$

The gradient product of two slant lines perpendicular to each other should equal to $- 1$, that is: they form a pair of negative reciprocals.

Hence, the slope of the normal would equal to the opposite of the slope of the tangent inversed

"Slope of Normal"=-1/("Slope of Tangent")

$\text{ } \textcolor{w h i t e}{l} = - \frac{1}{\frac{1}{2} \cdot \left(12 - \sqrt{3}\right)}$

Simplifying

$- \frac{1}{\frac{1}{2} \cdot \left(12 - \sqrt{3}\right)}$
$= 2 \cdot \frac{1}{\sqrt{3} - 12} \cdot \textcolor{g r e y}{\frac{\sqrt{3} + 12}{\sqrt{3} + 12}}$
$= - \frac{24 + 2 \cdot \sqrt{3}}{141}$