# What is the equation of the line normal to f(x)= sqrt(2x^3-x)  at x=1?

Feb 19, 2016

$y = \frac{5}{2} x - \frac{5}{2}$

#### Explanation:

Let $y = \sqrt{2 {x}^{3} - x}$

$y = {\left(2 {x}^{3} - x\right)}^{\frac{1}{2}}$

Applying the chain rule $\Rightarrow$

$y ' = \frac{1}{2} {\left(2 {x}^{3} - x\right)}^{- \frac{1}{2}} \times \left(6 {x}^{2} - 1\right)$

$y ' = \frac{\left(6 {x}^{2} - 1\right)}{2 \sqrt{2 {x}^{3} - x}}$

This equals the gradient $m$.

So when $x = 1 \Rightarrow$

$m = \frac{\left(6 - 1\right)}{2 \sqrt{2 - 1}} = \frac{5}{2}$

The tangent line is of the form:

$y = m x + c$

At $x = 1$,

$y = \sqrt{2 - 1} = 1$

$\therefore 1 = \frac{5}{2} \times 1 + c$

$\therefore c = - \frac{5}{2}$

So the equation of the tangent is:

$y = \frac{5}{2} x - \frac{5}{2}$