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)#.