Question

Let f be the functio given by f(x)=2x/sqr.(x^2+x+1)?

Answer 1

a.) Find the domain of f. Justify your answers
b.) In the viewing widow provided below, sketch the graph of f.
Viewing Window [-5,5] x [-3,3]
c.) Write an equation for each horizontal asymptote of the graph of f.
d.) Find the range of f. Use f'(x) to justify your answer
Note: f'(x)=(x+2)/(x^2+x+1)^(3/2)

Answer 2

since \(x^2+x+1>0\) for all \(x\) the domain is all real numbers

Answer 3

so how do i find domain?

Answer 4

you cannot take the square root of a negative number and \(\sqrt{x^2+x+1}\) is in the denominator, so it cannot be 0 as well
but \(x^2+x+1>0\) always, so you don't have to worry about that

Answer 5

you can check this because discriminant of \(x^2+x+1\) is negative

Answer 6

ok so whats the domain?

Answer 7

all real numbers

Answer 8

@satellite73 ok so for part b.) i just type in the function f(x)=2x/sqr.(x^2+x+1) on my calculator?

Answer 9

@satellite73 how do i do part b.)?

Answer 10

@satellite73 for part b.) i just type in the function f(x)=2x/sqr.(x^2+x+1) on my calculator?