# Question #123c3

Oct 5, 2017

Question 1: (I'm not sure what exactly the question wants to find)

Question 2: $\left(3 , \frac{3}{2}\right)$ and $\left(- 1 , \frac{1}{2}\right)$

#### Explanation:

I'm not quite sure what Question 1 is asking, but here's how you do Question 2:

Find the points on the curve of $f \left(x\right) = \frac{x}{x - 1}$ where the tangent line is parallel to the line $x + 4 y = 1$.

So, the question has indirectly given you the gradient of the tangent by telling you what the tangent line is parallel to.

By rearranging $x + 4 y = 1$, you can see that the gradient $= - \frac{1}{4}$.

To find where the curve has gradient $- \frac{1}{4}$, you then need to differentiate the function, using the quotient rule:

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

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

You can see that $u = x$, and $v = x - 1$

Let's differentiate these first, before subbing them into the formula.

$u = x$
$\therefore u ' = 1$

$v = x - 1$
$\therefore v ' = 1$

Subbing these into the formula, we get:

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

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

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

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

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

We know that the gradient we're looking for is $- \frac{1}{4}$.
When $y ' = - \frac{1}{4}$,

$- \frac{1}{4} = - \frac{1}{x - 1} ^ 2$

$4 = {\left(x - 1\right)}^{2}$

$x - 1 = \pm 2$

$\therefore x = 1 \pm - 2$

$\therefore x = - 1 , 3$

We now have the x-coordinates of the points at which the gradient is $- \frac{1}{4}$, and thus where the tangent lines would be parallel to $x + 4 y = 1$.

To find the y-coordinates, you just need to sub the points we just found back into the first function.

When $x = 3$,

$y = \frac{3}{3 - 1}$

$\therefore y = \frac{3}{2}$

And hence we get the point $\left(3 , \frac{3}{2}\right)$.

When $x = - 1$,

$y = \frac{- 1}{- 1 - 1}$

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

$\therefore y = \frac{1}{2}$

And hence we get the point $\left(- 1 , \frac{1}{2}\right)$.

In short, the answer to Question 2 is $\left(3 , \frac{3}{2}\right)$ and $\left(- 1 , \frac{1}{2}\right)$.