Question

Question #123c3

Answer 1

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

Question 2: `(3,3/2)` and `(-1,1/2)`

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(x)=x/(x-1)` where the tangent line is parallel to the line `x+4y=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+4y=1`, you can see that the gradient `=-1/4`.

To find where the curve has gradient `-1/4`, you then need to differentiate the function, using the quotient rule:

`f(x)=x/(x-1)`

`f'(x)=(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'(x)=(u'v-v'u)/v^2`

`f'(x)=(x'(x-1)-(x-1)'x)/(x-1)^2`

`f'(x)=(1(x-1)-1(x))/(x-1)^2`

`f'(x)=(x-1-x)/(x-1)^2`

`therefore f'(x)=-1/(x-1)^2`

We know that the gradient we're looking for is `-1/4`.
When `y'=-1/4`,

`-1/4=-1/(x-1)^2`

`4=(x-1)^2`

`x-1=+-2`

`therefore x=1+--2`

`therefore x=-1, 3`

We now have the x-coordinates of the points at which the gradient is `-1/4`, and thus where the tangent lines would be parallel to `x+4y=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=3/(3-1)`

`therefore y=3/(2)`

And hence we get the point `(3,3/2)`.

When `x=-1`,

`y=(-1)/(-1-1)`

`y=(-1)/(-2)`

`therefore y=1/2`

And hence we get the point `(-1,1/2)`.

In short, the answer to Question 2 is `(3,3/2)` and `(-1,1/2)`.