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

Jul 1, 2017

$y = - \frac{158}{9} \cdot x + \frac{211}{3}$

#### Explanation:

Let

$u = 2 {x}^{3} - 1$

$v = {x}^{2} - 2 x$

Then

$f \left(x\right) = \frac{u}{v}$

Using the quotient rule, which can be found in all good google searches:

$f ' \left(x\right) = \frac{u ' v - u v '}{v} ^ 2$

$u ' = 6 x$

$v ' = 2 x - 2$

$f ' \left(x\right) = \frac{\left[6 x \left({x}^{2} - 2 x\right)\right] - \left[\left(2 {x}^{3} - 1\right) \left(2 x - 2\right)\right]}{{x}^{2} - 2 x} ^ 2$

Soz, I cannot be bothered simplifying this...

$f ' \left(x\right)$ is the gradient function, substituting in any x value will give us the gradient of the tangent at that point.

So let's sub in x=3:

$f ' \left(3\right) = \frac{\left[6 \cdot 3 \left({3}^{2} - 2 \cdot 3\right)\right] - \left[\left(2 \cdot {3}^{3} - 1\right) \left(2 \cdot 3 - 2\right)\right]}{{3}^{2} - 2 \cdot 3} ^ 2$

$\Rightarrow f ' \left(3\right) = \frac{\left[18 \left(9 - 6\right)\right] - \left[\left(54 - 1\right) \left(6 - 2\right)\right]}{9 - 6} ^ 2 = \frac{18 \cdot 3 - 53 \cdot 4}{9} = - \frac{158}{9}$

Now we want the equation of the tangent, which is a straight line of the form:

$y = m x + c$

We know $m = - \frac{158}{9}$ but we need to solve for $c$. We need a point on the line to solve it, so sub in $x = 3$ into the original equation to find the y value:

$f \left(3\right) = \frac{2 \cdot {3}^{3} - 1}{{3}^{2} - 2 \cdot 3} = \frac{54 - 1}{9 - 6} = \frac{53}{3}$

So the coordinates of our point on the line are (3,53/3). Sub this point into the straight line equation to solve for c:

$y = - \frac{158}{9} \cdot x + c$

$\Rightarrow \frac{53}{3} = - \frac{158}{9} \cdot 3 + c$

$\Rightarrow \frac{53}{3} = - \frac{158}{3} + c$

$\Rightarrow c = \frac{53}{3} + \frac{158}{3} = \frac{211}{3}$

So the equation of the tangent is:

$y = - \frac{158}{9} \cdot x + \frac{211}{3}$