Question #123c3

1 Answer
Oct 5, 2017

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).