# Thread: tangents with specific slopes...

1. ## tangents with specific slopes...

so how do i derive this problem?

are there any points on the curve y=(x/2)+1/(2x-4) where the slope is -3/2?
where do i start?

thx!

2. Originally Posted by cyberdx16
so how do i derive this problem?

are there any points on the curve y=(x/2)+1/(2x-4) where the slope is -3/2?
where do i start?

thx!
The derivative is,
y'=(1/2)-2(2x-4)^(-2)
You need that to be -3/2
Thus,
1/2-2(2x-4)^(-2)=-3/2
Thus,
-2(2x-4)^(-2)=-2
Thus,
(2x-4)^(-2)=1
Thus,
2x-4=1
Thus,
x=2.5

3. If you learned MVT here is an alternative.
---
Now there is a longer but more elegant way to show that this is true. It uses the awesome faboulus mean-value theorem (most important theorem in calculus).
---
I am going to find an interval [a,b] such that,
f(b)-f(a)
--------
b-a

Would be equal to -3/2

Of course we need to check, if f(x) is continous on [a,b] and differenciable on (a,b).

Let us use a convienat point say a=0.
Then,
f(0)=-1/4

Thus, the mean value of the interval is,
f(b)+1/4
---------
b

We need this to be equal to -3/2
If we multiply both sides by b we have,
f(b)+1/4=(-3/2)b
Thus,
b/2+1/(2b-4)+1/4=(-3/2)b
Thus,
1/(2b-4)+1/4=-2b
Thus,
1/(2b-4)+2b=-1/4
Thus,
(1+4b^2-8b)/(2b-4)=-1/4
Thus,
4b^2-8b+1=1-b/2
Thus,
8b^2-16b=-b
Thus,
8b^2-15b=0
Thus,
b(8b-15)=0
Thus,
b=0 makes equation true but we cannot use it since b>a
Or,
b=15/8
Which makes the equation true.
Keeps the function continous on [0,15/8]
And differenciable on (0,15/8)

This mean that, on the interval (0,15/8)
there exists a point c such that,
f'(c)=MEAN VALUE=-3/2
Q.E.D.

4. (2x-4)^(-2)=1
Thus,
2x-4=1
Thus,
x=2.5[/QUOTE]

so to get rid of the (2x-4)^(-2)=1
did you squard both sides to get rid of the ^-2

EDIT: or i mean from (2x-4)^(-2)=1 to x=2.5 you would multiply by (2x-4)^(-2) which you would get 1=(2x-4)^2 and then sq both sides and set everything to x?

5. You have,
(2x-4)^(-2)=1
Thus,
Code:
1
----                 =1
(2x-4)^2
Thus,
multiply both sides by (2x-4)^2
1=(2x-4)^2
Just take the positive root cuz you are only trying to find one value that makes it true.
2x-4=1
-------------
The mental trick I do it take reciprocal of both sides.

6. Originally Posted by cyberdx16
so how do i derive this problem?

are there any points on the curve y=(x/2)+1/(2x-4) where the slope is -3/2?
where do i start?

thx!
Slightly generalised:

You are asked if there are points on the curve y=f(x), where the slope is m

The slope of the curve at x is dy/dx, so the question becomes does the
equation:

dy/dx=m

have any real solutions.

In your case f(x)=(x/2)+1/(2x-4), and m=-3/2, so we want to know if:

d/dx[(x/2)+1/(2x-4)]=-3/2

has any solutions.

The detail of answering this have been dealt with by the other posters.

RonL