algebra equation help (1 Viewer)

[abdooooo!!!]

x^2 - 3x + 4/x - 3, find the range

i saw this equation in a newspaper or something, it solved for y the range. the answer is y < and = -1, y > and = 7. but i can't understand the working out.

how the hell do you solve that? can someone explain???

 

[abdooooo!!!]

anyone?

 

[wogboy]

What about when x = 0.5

x^2 - 3x + 4/x - 3

= 0.25 - 1.5 + 8 - 3
= 3.75

which is out of the given range, so there's a mistake somewhere.

 

[abdooooo!!!]

Originally posted by wogboy
What about when x = 0.5

x^2 - 3x + 4/x - 3

= 0.25 - 1.5 + 8 - 3
= 3.75

which is out of the given range, so there's a mistake somewhere.

sorry my mistake... its x^2 - 3x +4/(x - 3) .

 

[PG5]

What about when x=-1 this time:

x^2 - 3x + 4/(x-3)

= 1 + 3 - 4/4
= 1 + 3 - 1
= 3

which is again out of the range you stated. Another mistake perhaps? Because if you sketch this equation, the range is all real y.

 

[Harimau]

Let y=x^2 - 3x + 4/(x-3)

Is a modified hyperbola with the vertical aymptote of x=3 and a curvic asymptote of (x^2 - 3x) hmmm... Damn this is a hard question... i tried subbing in y= -1 and and y=7 but that doesnt really work.

 

[wogboy]

It's wrong, the range of that function is all real y, here's the proof:

y=x^2 - 3x + 4/(x-3)

the limit of y as x approaches negative infinity is,

lim(x->-infinity) of (y)
= lim(x -> -infinity) of (x^2 - 3x + 4/(x-3))
= lim(x ->-infinity) of (x^2 - 3x) + lim(x->-infinity) of (4/(x-3)
= +infinity + 0
= +infinity

so lim(x->-infinity) of (y) = +infinity,

So the maximum value of y is infinity (i.e. there is no maximum).

Now let's look for lowest possible y value:

lim(x->3-) of (y)
= lim(x->3-) of (x^2 - 3x) + lim(x->3-) of (4/(3-x))
= 0 - infinity
= - infinity,

in other words the minumum value of y is - infinity. (i.e. there is no minumum),

and it's obvious that the only discontinuity in the graph of the function which is at x = 3, so between x = 3 (not included) and x = -infinity (i.e. for all x<3), y is continous. This means that in this domain of x-values (x<3), y can take on any value between minus infinity and plus infinity, or in other words any real value of y.

So the range of that function is all real y.

P.S. What's this Q doing in the 2U forum, it belongs in the 4U section?

 

[PG5]

There should be a minimum value though. Since x^2-3x is the curvic asymptote and x=3 is the vertical asymptote, f(x) approaches these two asymptotes but never touches them.

If you sub in x=4 into f(x), the y value would be:
= 4^2 - 3*4 + 4/(4-3)
= 16 - 12 + 4
= 8

for the equation x^2-3x, sub in x=4
= 16 - 12
= 4

Therefore:
for x>3, f(x) here is sketched b/w the two equations x^2-3x and x=3 as they are both asymptotes of f(x) and the f(x) value of x=4 is above the curvic asymptote x^2-3x. ie. it should turn out to be a concave up parabola in b/w those equations and thus having a minimum value.

 

[wogboy]

That's right, but this minumum point you're referring to is actually a RELATIVE minima, not a absolute minima. However, if you look to the left of the asymptote x = 3, you will find that there are no minima nor maxima here and all values of y are catered for. For example take x to be a number just smaller than 3 like 2.99.

When x = 2.99,

y = (2.99)^2 - 3(2.99) + 4/(3-2.99)
~ - 400

Then take for instance x = 2.999

y = (2.999)^2 - 3(2.999) + 4/(3 - 2.999)
~ -4000

which are very large negative numbers. And as x gets close to 3, the graph shoots right down to minus infinity, as a vertical asymptote.

So while there is a relative minima somewhere at x>3, there is no absolute minima, and the range is all real y.