1. ## Coordinate Geometry

Hello again! I'm in need of some help with these two parts.

Find the value(s) of k for which the equation x + 1 + (1/x) - k = 0, x not equal to 0, has no real solutions.

A second function is defined as g(x) = b/(f(x)), b greater than 0, where f(x) = (x^2 + x + 1)/(x), x not equal to 0. Determine the range of g(x), leaving the answer in terms of b, and clearly explain your reasoning.

2. For the first one.
Because x is not zero, multiply through by x: x^2 +(1-k)x +1 =0.
Make the discriminant less than zero.

3. Thanks for the help, still stuck on the 2nd part.

4. Originally Posted by classicstrings
Hello again! I'm in need of some help with these two parts.

Find the value(s) of k for which the equation x + 1 + (1/x) - k = 0, x not equal to 0, has no real solutions.

A second function is defined as g(x) = b/(f(x)), b greater than 0, where f(x) = (x^2 + x + 1)/(x), x not equal to 0. Determine the range of g(x), leaving the answer in terms of b, and clearly explain your reasoning.
For the first problem, I would subtract 1/x then multiply both sides by x:
x(x + 1 - k) = -1
x^2 + (1 - k)x + 1 = 0

x = {-(1 - k) +/- sqrt[(1 - k)^2 - 4]}/2

Now we need to find where this will have no real values, which means we only need to see where sqrt[(1 - k)^2 - 4] does not exist, thus were (1 - k)^2 - 4 < 0.

(1 - k)^2 - 4 < 0
(1 - k)^2 < 4
|1 - k| < 2

This gives us two sets of problems:
1 - k < 2
k > -1

1 - k > -2
k < 3

Therefore, for k is any value in the range: (-1,3), the function above has no real roots.

5. Originally Posted by classicstrings
Hello again! I'm in need of some help with these two parts.

Find the value(s) of k for which the equation x + 1 + (1/x) - k = 0, x not equal to 0, has no real solutions.

A second function is defined as g(x) = b/(f(x)), b greater than 0, where f(x) = (x^2 + x + 1)/(x), x not equal to 0. Determine the range of g(x), leaving the answer in terms of b, and clearly explain your reasoning.
g(x) = (bx)/(x^2 + x + 1) for x != 0 and b > 0.

Well, to find it's range, we need to find it max/min values. Take it's derivative and set it equal to 0 and also determine where it's undefined. Determine if it's increasing/decreasing on either side of this critical point and find the value of the function at this critical point. I think I can safely guarantee that range will be (-inf,0)U(0,inf) no matter what b is, but that's something you'll have to check.

6. I get the derivative to be
EDIT - made error in calculation.

But the question wants the range in terms of b.

7. Originally Posted by classicstrings
I get the derivative to be

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

But the question wants the range in terms of b.
g(x) = (bx)/(x^2 + x + 1)
g'(x) = [b(x^2 + x + 1) - bx(2x + 1)]/(x^2 + x + 1)^2
g'(x) = [bx^2 + bx + b - 2bx^2 - bx]/(x^2 + x + 1)^2
g'(x) = [-bx^2 + b]/(x^2 + x + 1)^2

Let g'(x) = 0:
-bx^2 + b = 0 --> x^2 = 1 --> x = 1, -1

Let g'(x) be undefined:
(x^2 + x + 1)^2 = 0
x^2 + x + 1 = 0
This results in imaginary solutions, so we don't need to deal with this.

Now using the points x = 1, -1 in g(x) we get:
g(1) = b/3
g(-1) = -b

Now, we need to see if g is increasing/decreasing on the intervals (-inf,-1), (-1,0), (0,1), (1,inf).

In g'(x):
(-inf,-1) --> g'(x) --> {negative}
(-1,0) --> g'(x) --> {positive}
(0,1) --> g'(x) --> {positive}
(1,inf) --> g'(x) --> {negative}

x = -1 is a minimum
x = 1 is a maximum

My conclusions after this point are wrong.
------------------------------------------------------------------------------------

Therefore, there are two parts to this graph, one is always above g(-1) = b/3, the other is always below g(1) = -b.

The range is: (-inf,-b) U (b/3,inf)

8. It appears I was wrong about my original prediction that the range would be from (-inf,0) U (0,inf).

9. Hello, classicstrings!

A second function is defined as: .g(x) .= .b/(f(x)), .b > 0,
. . where f(x) .= .(x² + x + 1)/(x), .x ≠0.

Determine the range of g(x), leaving the answer in terms of b,
I will assume this is a Calculus problem and derivatives are allowed.

We have: .g(x) .= .bx/(x² + x + 1)

. . . . . . . . . . . . . . . . . . . .b(x² + x + 1) - bx(2x + 1)
The derivative is: .g'(x) .= .--------------------------------
. . . . . . . . . . . . . . . . . . . . . . . (x² + x + 1)²

. . . . . . . . . . . . . . . . . . . . . . . . -b(x² - 1)
. . which simplifies to: .g'(x) .= .---------------- . = . 0
. . . . . . . . . . . . . . . . . . . . . . .(x² + x + 1)²

. . and has critical values: .x .= .±1

The corresponding y-values are: .y .= .b/3, -b

We see that the x-axis is a horizontal asymptote.

Hence: (1, b/3) is a maximum of the graph
. . and: (-1, -b) is a minimum of the graph.

The range of the function is: . -b .< .g(x) .< .b/3

10. Thanks Soroban,

I'm still a bit blurry, how do you realise that y = 0 is an asymptote and deduce that (1,b/3) is the max and (-1,-b) is min?

-> For f(x) = (x^2 + x + 1)/(x), the asymptotes are y = x +1 and x = 0 right?

11. Originally Posted by classicstrings
Thanks Soroban,

I'm still a bit blurry, how do you realise that y = 0 is an asymptote and deduce that (1,b/3) is the max and (-1,-b) is min?

-> For f(x) = (x^2 + x + 1)/(x), the asymptotes are y = x +1 and x = 0 right?
From g(x) = bx/(x^2 + x + 1)

Lim{x->inf} g(x) = 0
Lim{x->-inf} g(x) = 0

It's the g(x) function from which we determine the horizontal asymptotes.

And if you look back at the work I did, even though my conclusion was wrong (I did it quickly just before having to head off to school), I showed that at x = -1, g(x) has a relative minimum, at x = 1, g(x) has a relative maximum.