If you learned MVT here is an alternative.
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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).
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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.
Nice thing about this is that it is an existence theorem which I love.
(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?
You have,
(2x-4)^(-2)=1
Thus,
Thus,Code:1 ---- =1 (2x-4)^2
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
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The mental trick I do it take reciprocal of both sides.
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