How do you find the domain and range of #f(x) = (x) / sqrt(x^2+x+1)#?

Answer 1

Domain: #(-oo, oo)#

Range: #[-2/3sqrt(3), 1)#

Take note of this:

#x^2+x+1 = (x+1/2)^2+3/4 > 0" "# for all real values of #x#
So #sqrt(x^2+x+1)# is well defined and non-zero for any #x in RR#
Hence #f(x)# is well defined for any #x in RR#
So its implicit domain is #RR#

Find the range by letting:

#y = f(x) = x/sqrt(x^2+x+1)#

Next:

#y sqrt(x^2+x+1) = x#

Taking both sides equal:

#y^2(x^2+x+1) = x^2#

Thus:

#(y^2-1)x^2+y^2x+y^2=0#
This is a quadratic polynomial equation in #x# of the form:
#ax^2+bx+c = 0#

along with

#{ (a = y^2-1), (b=y^2), (c=y^2) :}#
Its discriminant #Delta# is given by the formula:
#Delta = b^2-4ac#
#color(white)(Delta) = (y^2)^2-4(y^2-1)y^2#
#color(white)(Delta) = y^4-4y^4+4y^2#
#color(white)(Delta) = y^2(4-3y^2)#
When #y=0# then #Delta=0# so the quadratic in #x# has one solution.
Otherwise #y^2 > 0# so in order for the quadratic in #x# to have real solutions we require:
#4-3y^2 >= 0#

That is:

#4 >= 3y^2#

Thus:

#y^2 <= 4/3#

additionally:

#abs(y) <= sqrt(4/3) = sqrt((4*3)/9) = 2/3sqrt(3)#
This is a sufficient condiction for the polynomial to have a solution, but we also require the sign of #x# to match that of #y#. This is where we run into the limitation of squaring an equation that we did earlier - which can introduce spurious solutions.
Note that in order to have positive solutions, the signs of the coefficients #a#, #b# and #c# must differ. Hence we need #y^2-1 < 0#. So #abs(y) < 1#.
So the range of #f(x)# is #[-2/3sqrt(3), 1)#

plot{x/sqrt(x^2+x+1) [-5, 5, -2.5, 2.5]}

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Answer 2

Domain: #(-oo, oo)#

Range: #[-2/3sqrt(3), 1)#

This approach makes use of some calculus and pre-calculus.

Note that #x^2+x+1 = (x+1/2)^2+3/4 > 0# for all real values of #x#
Hence for any #x in RR# we have:
#sqrt(x^2+x+1) > 0#
Hence #f(x)# is well defined for all #x in RR#
So the domain of #f(x)# is #RR#

Take note of this:

#lim_(x->-oo) x/sqrt(x^2+x+1) = lim_(x->-oo) -1/sqrt(1+1/x+1/x^2) = -1#
#lim_(x->oo) x/sqrt(x^2+x+1) = lim_(x->oo) 1/sqrt(1+1/x+1/x^2) = 1#
So there are horizontal asymptotes #y=-1# as #x->-oo# and #y=1# as #x->oo#.
#d/(dx) x/sqrt(x^2+x+1) = d/(dx) x (x^2+x+1)^(-1/2)#
#color(white)(d/(dx) x/sqrt(x^2+x+1)) = (x^2+x+1)^(-1/2) - x 1/2(x^2+x+1)^(-3/2)(2x+1)#
#color(white)(d/(dx) x/sqrt(x^2+x+1)) = (x^2+x+1)^(-3/2)((x^2+x+1) - 1/2x(2x+1))#
#color(white)(d/(dx) x/sqrt(x^2+x+1)) = (x+2)/(2(x^2+x+1)^(3/2))#
So there's one turning point at #x=-2#

We discover:

#f(-2) = (-2)/sqrt((-2)^2+(-2)+1) = -2/sqrt(3) = -2/3sqrt(3)#
Hence the range of #f(x)# is:
#[-2/3sqrt(3), 1)#

plot{x/sqrt(x^2+x+1) [-5, 5, -2.5, 2.5]}

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Answer 3

To find the domain of the function ( f(x) = \frac{x}{\sqrt{x^2+x+1}} ), we need to ensure that the expression under the square root, ( x^2+x+1 ), is positive. This is because the square root of a negative number is undefined in the real number system. Therefore, we need to solve the inequality ( x^2+x+1 > 0 ) to determine the values of ( x ) that are valid for the domain.

To find the range of the function, we need to consider the behavior of the function as ( x ) approaches positive and negative infinity. Additionally, since the function is continuous and the denominator is never zero, there are no vertical asymptotes. Therefore, the range of the function is all real numbers.

In summary:

  • Domain: ( x^2+x+1 > 0 )
  • Range: All real numbers
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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