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.
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
Using the quadratic equation:
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.
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.
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)
Hello, classicstrings!
I will assume this is a Calculus problem and derivatives are allowed.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,
and clearly explain your reasoning.
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
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.