5 algebra problems!!!!!?
1) find all zeros (rational, irrational & imaginary)
P(x)= x^4+5x^2+6
2) give an example of a rational function that satisfies these conditions:
real zeros: none; vertical asymptotes: none; horizontal asymptote: 3
3) The total cost of producing x units of a certain product is given by:
C(x)= (1/5)x^2+2x+2000
The average cost per unit for producing x units is D(x)=C(x)/x
a) find the rational function D(x)
b) at what production level will the average cost per unit be minimal?
4) find all solutions
3xe^(-x)+x^(2)e^(-x)=0
5) solve
ln(x+1)=ln(3x+1)-ln(x)
problems 4 and 5 are the most important!!!!!
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1) find all zeros (rational, irrational & imaginary)
P(x)= x^4+5x^2+6
1) find all zeros (rational, irrational & imaginary)
P(x)= x^4+5x^2+6 = (x^2+2)(x^2+3).
So the solutions are the solutions to x^2+2 = 0 and x^3+3=0, and those are easy to solve.
2) I don't think a function can have 3 horizontal asymptotes. If it did, then it wouldn't be a function. (Try picking 3 horizontal asymptotes, draw them, and then try drawing a function that approaches them as it heads to +-infinity.)
Now, I haven't spent a lot of time thinking about this. Over the complex numbers I might be wrong.
3) Sorry, I didn't think about this one at all. It looks more like detailed work than hard thought.
4) 3xe^(-x)+x^(2)e^(-x) = (e^(-x))(x(x+3)).
So the zeroes are x=0,-3.
I don't think e^{-x} can ever be zero.
Taking natural logs would lead to an impossibility.
5) ln(x+1)=ln(3x+1)-ln(x)
Exponentiation undoes ln(), so raise e to each side to get (x+1) = (3x+1)/x.
From there it's just solving a quadratic equation.
I hope this has helped.