Prove using the algebraic method, that f(x)>=a or f(x)<=b, where a and b are exact values to be determined

Question

Consider the curve y=f(x), where f(x)= (x^2-1)/(x+2). Prove using the algebraic method, that f(x)>=a or f(x)<=b, where a and b are exact values to be determined.

Can you show me what is the so called algebraic method?

Answer 1

Lets see what values we can have for f(x). Suppose f(x) = c has some solution for x.
Then (x^2 - 1) = c(x+2), so x^2 - cx - (2c+1) = 0. For that to have a solution, we need the discriminant, (b^2-4ac, though note thats a different c!) to be at least 0. So we need c^2+4(2c+1) >= 0.
So c^2 + 8c + 1 >= 0.
Now, we can write that as (c+4)^2 - 15 >= 0, so (c+4)^2 >= 15.
Thus c+4 > sqrt(15) or c+4 < -sqrt(15).
So c > sqrt(15)-4 or c <-sqrt(15)-4.

So f(x) > sqrt(15)-4 or f(x) < -sqrt(15) - 4.

(Note you could have done the last step solving the 'c' inequality with the quadratic formula; they're both the same, really.)

Note that it is a much better idea to leave the inequality until as late as possible. Note that in the above idea of gp4rts, although it may be made to work, you *can't* do the following:
(x^2-1)/(x+2) >= a
Therefore
(x^2-1) >= a(x+2).
If you multiply by a negative, you have to switch the inequality. Thus you would have to divide everything into two cases - if x+2>0, or x+2<0, and it gets very messy.

Answer 2

f(x)= (x² - 1)/(x + 2)
= (x² - 4 + 3)/(x + 2)
= (x² - 4)/(x + 2) + 3/(x + 2)
= (x - 2)(x + 2)/(x + 2) + 3/(x + 2)
= x - 2 + 3/(x + 2)

1. There is a discontinuity at x = -2 as x + 2 = 0 so 3/(x + 2) is undefined. Thus x = -2 is a vertical asymptote.

2. When x = 0 f(x) = -1/2 So it cuts the y-axis at x = -1/2

3. When x increases and decreases without bound 3/(x + 2) approaches 0 and so f(x) approaches the line f(x) = x - 2
ie it is asymptotic to the line f(x) = x -2

4. When x < -2, 3/(x + 2) < 0. So f(x) = x - 2 + 3/(x + 2) < x - 2
So when x < -2 there must be a maximum value of f(x) = x - 2 + 3/(x + 2)

By calculus you can show that this occurs when x = -2 - √3 (I do not know how to get this value algebraically)
and when this occurs f(-2 - √3) = -4 - 2√3
So when x < -2, f(x) ≤ -4 - 2√3

When x > -2, 3/(x + 2) 0. So f(x) = x - 2 + 3/(x + 2) > x - 2
So when x > -2 there must be a maximum value of f(x) = x - 2 + 3/(x + 2)

By calculus you can show that this occurs when x = -2 + √3 (I do not know how to get this value algebraically)
and when this occurs f(-2 + √3) = -4 + 2√3
So when x > 2 f(x) ≥ -4 + 2√3

Answer 3

I assume using algebraic method means solving this using algebra. To do this you write an equation that expresses what you are after and then manipulate it algebraically to get an answer. In the first case, you want

f(x) ≥ a

write it out, using the expression you are given for f(x)

(x^2-1)/(x+2) ≥ a

multiply both sides by (x-2); this does not change in inequality

x^2 - 1 ≥ a*(x + 2)

expand the right side

x^2 - 1 ≥ a*x + 2*a

move all terms to the left to get

x^2 - a*x -1 - 2*a ≥ 0

Solve for a. Check the result for close values around the solution for any given value of x to make sure you have the x values that make f(x) ≥ a.

If you take a = 4, for example, the solution to the quadratic gives you x = 5.606 and x = -1.606, If you plug the first value into the original equation for f(x) you get f(5.606) = 4; set x = 5.607 and you get f(5.607)= 4.001. Therefore, f(x) ≥ 4 for x ≥ 5.606. (The other value comes out -4 and doesn't fit the requirements.)

Solve the other the same way. All of these are algebraic steps with no calculus involved.

Answer 4

using alegraibic method(think so..):

y= (x^2-1)/(x+2)
dy/dx = (2x(x+2) - (x^2-1))/(x+2)^2
dy/dx = (2x^2+4x-x^2+1)/(x+2)^2
= (x^2+4x+1)/(x+2)^2

so make dy/dx = 0 to find max and mins (that's what being asked in the question)
x^2+4x+1=0

find zeros using quadratic formula:
x= -2 (plus minus thing) underoot (3)
or = -2+(3)^0.5 or -2-(3)^0.5
plug this in the main formaula:

and then those would be ur maxs and mins


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Editing..
think i have got the right answer, and it's through calculus method...

"By calculus you can show that this occurs when x = -2 + √3 (I do not know how to get this value algebraically)
and when this occurs f(-2 + √3) = -4 + 2√3
So when x > 2 f(x) ≥ -4 + 2√3"
Walc C

waow.. that was something like inutiative.

Answer 5

This is the method!
[1. y>=a for a>=0; (x^2-1)/(x+2)>=a
[1a. if x^2-1>0 & x+2>0, then x^2-1>=a*x+2a or x^2-ax-(2a+1)>=0
x1=[a-sqrt(a^2+8a+4)]/2, x2=[a+sqrt(a^2+8a+4)]/2; sqrt exists as a>=0;
thus if (x-a+r)*(x-a-r)>=0
it either means that x>=x1 & x>=x2. it is true for x>=x2
or means that x<=x1 & x<=x2. it is true for x<=x1.
[1b. (still y>=a for a>=0; (x^2-1)/(x+2)>=a)
if x^2-1<=0 & x+2<=0, then x^2-1<=a*x+2a
thus if (x-a+r)*(x-a-r)<=0, then x1<=x<=x2
[1c. if x^2-1<=0 & x+2>=0 … do it!
[1d. if x^2-1>=0 & x+2<=0 … do it!
[2. y>=a for a<=0; (x^2-1)/(x+2)>=a or y>=-|a|
[2.a,b,c,d … do it!
Mind the rule! if k/(-1)>m, then k<-m; if k/(-1)-m
Oh poor kids!!!

Answer 6

first graph the equation of f(x)= (x^2-1)/(x+2) and y=f(x) on your TI-83
if you want to do the problem algebraicly then you could plug in some numbers say for example let f(x)= 1,2,3,4,5,6 and so on

Answer 7

f(x)>=a

to find a, do derivative of f(x), set it equal to zero, find x and plug it back into f(.)