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

Jan 29, 2017

$y - \sqrt{3} = \frac{2 \sqrt{3}}{5} \left(x + 1\right)$

Explanation:

From the power rule, we see that $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$. The chain rule tells us that $\frac{d}{\mathrm{dx}} {u}^{n} = n {u}^{n - 1} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

So then:

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

Then:

$f ' \left(x\right) = \frac{1}{2} {\left(2 {x}^{2} - x\right)}^{- \frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left(2 {x}^{2} - x\right)$

Through the power rule we see that $\frac{d}{\mathrm{dx}} \left(2 {x}^{2} - x\right) = 4 x - 1$, so:

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

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

Now that we have the derivative, we need to write the equation of the normal line at $x = - 1$. To write the equation a line, we need to know two things: a point that the line passes through and the slope of the line.

The line will pass through the point on the curve $f$ at $x = - 1$.

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

So the normal line passes through the point $\left(- 1 , \sqrt{3}\right)$.

The normal line is perpendicular to the tangent line. The tangent line's slope is found through the value of the derivative at a point, which here is $f ' \left(- 1\right)$.

Perpendicular slopes are opposite reciprocals, so the slope of the normal line is $- \frac{1}{f ' \left(- 1\right)}$.

First:

$f ' \left(- 1\right) = \frac{4 \left(- 1\right) - 1}{2 \sqrt{2 {\left(- 1\right)}^{2} - \left(- 1\right)}} = \frac{- 5}{2 \sqrt{3}}$

Thus the slope of the normal line is:

$- \frac{1}{f ' \left(- 1\right)} = \frac{2 \sqrt{3}}{5}$

A line with slope $m$ passing through point $\left({x}_{1} , {y}_{1}\right)$ can be expressed as:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

The normal line has slope $\frac{2 \sqrt{3}}{5}$ and passes through $\left(- 1 , \sqrt{3}\right)$, so the normal line is:

$y - \sqrt{3} = \frac{2 \sqrt{3}}{5} \left(x + 1\right)$