A Chris Special
Find the maximum and minimum values of (x^2 -2x + 2)/(2x-2) without using Calculus!
posted by Zaux at
8:55 PM
Wednesday, February 10, 2010
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20 Comments:
what's Calculus?
Knightmare, Chris could provide a better definition or explanation, but Calculus is:
A branch on mathematics concerned with limits and the differentiation and integration of functions
... mmmmm, sounds pretty nasty when you write it out like that
Its -infinity and +infinity as there is no range specification
Its range is -infinity to +infinity, with the exception of +1. Hmm, what would be the max and minimum value?
The answer is -
∫[(x³ - 2x) / (2x² - 3x + 2)] dx
Kylee
Kylee, that looks suspiciously like Calculus!
Without using calculus, let's simplify the function a bit.
f(x) = (x^2-2x+2)/(2x-2)
=((2x-2)(x/2-1) + x)/(2x-2)
=(x/2-1) + x/(2x-2)
=x/2[1+1/(x-1)] - 1
Let's look at all the ranges of x to work out the limits of f(x).
By inspection, at x=2, f(x)=1. Increasing x from this point continues to increase f(x) so the upper limit of f(x) is infinity.
Decreasing x towards 1 will cause f(x) to approach infinity again, as 1/(x-1) will approach infinity.
Again by inspection, at x=0, f(x)=-1. Decreasing x from this point will continue to decrease f(x) so the lower limit of f(x) is negative infinity.
Increasing x from 0 towards 1 also causes f(x) to approach negative infinity as 1/(x-1) will approach negative infinity.
So the limits of the function are plus and minus infinity, but the range of the function is only from 1 to infinity and from -1 to -infinity, ie there are no values of x where -1<f(x)<1
According to the published source, there are numeric solutions.
(x^2-2x+2)=(x-1)(x-1)+1
2x-2=2(x-1)
eq=((x-1)(x-1)+1)/(2(x-1))
eq=1/2*((x-1)+1/(x-1))
if x>>1 and x-> inf (1/2*inf+1/inf)-->inf+0, thus eq->inf
if x<<1 and x-> -inf (1/2*-inf+1/-inf)-->-inf-0, thus eq-> -inf
now if x>1 and x->1, 1/2*((x-1)+1/(x-1)),1/2*(0+1/(0))-->0+inf-> inf
now if x<1 and x->1, 1/2*((x-1)+1/(x-1)),1/2*(0+1/(-0))-->0-inf->- inf
1/0 is not a number (NaN) but the above is to illustrate
Answer:
min values -inf at x=-inf and x=1 approach from x<1
max value +inf a x=inf and x=1 approach from x>1
but as noted at x=1 the function is undefined.
Cam
If source provides numeric solutions for max or min, possibly they are for the local minima, local maxima, just on either side of x=1.
As this function is the equation of a line y=1/2*x at any significant distance away from x=1. And lines with a slope !=0 will have a max of inf and min of -inf.
Plot of function confirms.
Cam
Local minima at x>1 is at x=2
for y=1
Local maxima at x<1 is at x=0 for y=-1
Cam
The eq=1/2*((x-1)+1/(x-1))
equation is simply (1/2)of y=1/u+u
where u=x-1 i.e. graph shifted one unit to the right
1/u local minima with u>0 is u=1
why ?
if 0>u<1 then 1/u >1
if u>1 then 1/u <1
but u=x-1 so x=2 is local minima with y=1
1/u local maxima with u< is u=-1
why ?
if -1>u<0 then 1/u <-1
if u<-1 then 1/u >-1
but u=x-1 so x=0 is local minima with y=-1
Cam
correction for last post...
"so x=0 is local minima with y=-1"
should be
"so x=0 is local maxima with y=-1"
Cam
∫(x)= (x²-2x+2)/(2x-x)=
1/2[x-1+(1/x-1)]=
x²/2(x-1)-1
The smallest absolute values of the sum of a number and its reciprocal are attained when the number is ±1. Hence the smallest aabsolute values of ∫(x) will be given by x-1 = ±1, so x = 2 or 0.
For the last form of ∫(x), it is evident that ∫(x) will increase without bound for x > 2, will increase negativelywithout bound for x < 0, and is undefined for
x = 1.
It follows that for x = 2, ∫(x) has a relative minimum value of 1, and for x = 0, ∫(x) has a relative maximum value of -1.
This post has been removed by the author.
Cam is the grand prize winner of ... mmmm ... well, congrats :-)
And Chris is a co-winner :-)
Let y = (x^2-2x+2)/(2x-2)
See the plot via wolframalpha: plot
Manually. Let x -> ∞, then y -> x/2, a straight line, slope 1/2
through the origin. y-> -∞ as x -> 1 from the left and y-> +∞ as
x -> 1 from the right.
From the plot can see that y has a local maxima at x ≈ 0
(then y ≈ -1) and a local minima at x ≈ 2 (then y ≈ 1).
But the graph has no solutions for -1 < y < 1 (approx).
Rearranging the equation => x² - 2x(1+y) + 2(1+y) = 0
The x = (2(1+y) +/- sqrt(4(1+y)²-8(1+y)))/2
X can only be real if (4(1+y)²-8(1+y) ≥ 0 and then then x = 1+y
Clearly the points of interest are when (1+y)² = 2(1+y)
i.e. x just real. That equation is satisfied if (1+y) = 0,
and then y = -1, or if (1+y) ≠ 0, then (1+y) = 2 => y = 1.
So minima at x,y = 0,-1 and maxima at x,y = 2,1.
I must have gone to bed when you posted. All the real activity seems to have happened in the last hour.
I deleted my first post as I'd got a significant typo in it.
My "and then x = 1+y" is too early.
... and I've swapped minima and maxima on my last line. More wake-up coffee required.
As the Frankenstein monster said in, I believe the first movie, "Smoke good" ... well, "Coffee good!" :-)
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