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 and what are the other non-algebraic methods of showing this?

HUH! I can not understnad anything

Answer 1

Algebraic methods refers to using symbol manipulation rather than calculus. The answer, if you use calculus, is a simple local max/min question.

Try to see when you can solve it:

(x^2 - 1)/(x + 2) = k

x^2 - 1 = xk + 2k

x^2 - xk - 2k - 1 = 0

so, using the quadratic formula,
x = (k +- sqrt(k^2 - 4(-2k - 1)))/2
= (k +- sqrt(k^2 + 8k + 4))/2
When are there no solutions to this? When k^2 + 8k + 4 < 0. So these are the values of k for which there are no real solutions for x. Solving with the quadratic equation, this is true for -4 +- sqrt(12), so the interval you are looking for is -4 - 2sqrt(3) < y < -4 + 2sqrt(3)

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

The function is defined everywhere except where x = -2. It has an asymptote there and goes to +oo and -oo.

In the limit for x -> +-oo, we have f(x)/x -> 1, so that again the function attains +- oo.

The local maximum and minimum values are found by equation the derivative with zero. It is
f'(x) = (2x(x+2) - (x^2-1)1)/(x+2)^2
= (2x^2 + 4x - x^2 + 1)/(x+2)^2
= (x^2 + 4x + 1)/(x^2 + 4x + 4)
f'(x) = 0 when the numerator is zero, so
... x^2 + 4x + 1 = 0
... (x + 2)^2 = 3
... x = -2 +- V3

Substitute back if f(x):
f(-2-V3) = (6 + 4V3)/(-V3) = -(2V3 + 4)
f(-2+V3) = (6 - 4V3)/V3 = 2V3 - 4

The first value is clearly the smallest and would qualify for b (local maximum); the second value is b (local minimum).

Answer 4

(x^2-1)/(x+2)

Limit (x^2-1)/(x+2) = limit x^2/x = limit x = +/- oo,
x->+/-oo................. x->+/-oo.......x->+/.oo

So, there are neither an upper nor a lower bound of f

Non algebraic methods could be graphic methods, computational methods, etc.

Ana