# What is the equation of the normal line of f(x)= (3x^2-2)/(6x) at x = 1?

Jan 4, 2016

$\textcolor{g r e e n}{\text{y=-6/5x+41/30}}$

#### Explanation:

$f \left(x\right) = \frac{3 {x}^{2} - 2}{6 x}$
Let us first find the slope of the tangent.
Slope of the tangent at a point is the first derivative of the curve at the point.
so First derivative of f(x) at x=1 is the slope of the tangent at x=1
To find f'(x) we need to use quotient rule
Quotient rule: $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{\frac{\mathrm{du}}{\mathrm{dx}} v - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
$u = 3 {x}^{2} - 2 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 6 x$
$v = 6 x \implies \frac{\mathrm{dv}}{\mathrm{dx}} = 6$
$f ' \left(x\right) = \frac{\frac{\mathrm{du}}{\mathrm{dx}} v - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
$f ' \left(x\right) = \frac{6 x \left(6 x\right) - \left(3 {x}^{2} - 2\right) 6}{6 x} ^ 2$
$f ' \left(x\right) = \frac{36 {x}^{2} - 18 {x}^{2} + 12}{6 x} ^ 2$$\textcolor{b l u e}{\text{combine the like terms}}$
$f ' \left(x\right) = \frac{18 {x}^{2} + 12}{36 {x}^{2}} \textcolor{b l u e}{\text{factor out 6 on the numerator }}$
$f ' \left(x\right) = \frac{6 \left(3 {x}^{2} + 2\right)}{36 {x}^{2}} \textcolor{b l u e}{\text{cancel the 6 with the 36 in the denominator}}$
$f ' \left(x\right) = \frac{3 {x}^{2} + 2}{6 {x}^{2}}$
$f ' \left(1\right) = \frac{3 + 2}{6} \implies f ' \left(1\right) = \frac{5}{6}$
$\textcolor{g r e e n}{\text{slope of the tangent=5/6}}$
$\textcolor{g r e e n}{\text{slope of the normal=negative reciprocal of slope of the tangent=-6/5}}$
$f \left(1\right) = \frac{3 - 2}{6} \implies f ' \left(1\right) = \frac{1}{6}$
$\textcolor{red}{\text{ point-slope form of an equation of line }}$
$\textcolor{red}{\text{y-y1=m(x-x1)...(where m:slope,(x1,y1) :points)}}$
We have slope =$- \frac{6}{5}$and the points are $\left(1 , \frac{1}{6}\right)$
Use the point slope form
$y - \left(\frac{1}{6}\right) = - \frac{6}{5} \left(x - 1\right) \implies y = \left(- \frac{6}{5}\right) x + \frac{6}{5} + \frac{1}{6}$
$\textcolor{g r e e n}{\text{combine the constant terms}}$
$\textcolor{g r e e n}{\text{y=-6/5x+41/30}}$