Consider the function f(x)=(3x^4-5x+3)/(x^4+1). Find each of the following (if they exist):?

Question

(a) domain
(b) vertical, horizontal, or slant asymptotes
(c) x-intercept(s)
(d) y-intercept(s)
(e) symmetry (with respect to the x-axis, y-axis or the origin?
(f) f'(x)
(g) critical numbers
(h) f"(x)
(i) possible points of inflection

Answer 1

In addition to mr. Zolex above.
Thus y(x) = 3 - 5x/(x^4+1); if x=0 then y=3; if y=0 then no real number for x; thus the whole y(x) is above X-axis (see below why). Besides y(x) has no symmetry.
No vertical asymptotes;
Now lim y(x) {for |x|->inf}=3 means that the line y=3 is bi-directional horizontal asymptote.
Let r=x^4+1, then y’=-{5x^4+5 –(4x^3)*(5x)} / r^2 =5{3x^4 -1}/r^2, x=m*(1/3)^(1/4) possible 2 local extreme points, m=-1 or m=+1.
Cusps? y’’=5{12x^3 * r - 2*(4x^3)* (3x^4-1)} / r^3 = 5{12x^7 + 12x^3 – 24x^7 + 8x^3} / r^3 = 20x^3{5-3x^4) / r^3, hence 3 cusps at x=0 and x=m*(5/3)^(1/4);
If x<-(5/3)^(1/4) then y’’>0 y is concave;
If –(5/3)^(1/4)0 y is concave; if x>(5/3)^(1/4) y’’<0 y is convex again.
If x1=-(1/3)^(1/4) then y’’<0 y(x1)=5.849384 is local and absolute max!
If x2=+(1/3)^(1/4) then y’’>0 y(x2)=0.150616 is local and absolute min!

Answer 2

f(x)=(3x^4-5x+3)/(x^4+1)
The denominator is always positive and will never = 0, therefore the domaine is all real values of x, and there will be no vertical asymptotes..

Since the numerator and denominator are of the same degree, there will be a horizontal asymptote at y-3, because 3x^4/x^4=3, and there will be no slant asymptotes.

Plotting a few values of x and y help to see the graph.This is most easily done on a spreadsheet if you don't have a calculator.

The graph is asymptotic to the li ne y = 3 from - infinity and risese to a maximum at somewhere between -.9 and -.8. It the n turns dowward passing through the y-axis at (0.3) and then reaches a minimum between x = .8 and .9 and then proceeds upward and asymptoically to th line y =3.

The function is concave upward from -infinity < x <0 and concave downward from 0
The point (0,3) is a point of inflection where concavity changes.

The function has no x-intercepts as its roots are all imaginary.

Critical points are between -,9 and -.8 where the local maximum occurs annd between ,8 and .9 where the local minimum occurs.

The function is symmetric about the point (0,3)

f'(x) = [(x^4+1)(4x^3-5) - (3x^4-5x+3)(4x^3)]/(x^4+1)^2I'll leave it to you to simplify.

The answer 0 part f, is f(x) which was given.

Answer 3

only positioned: virtually no human being receives with the help of existence without falling out of grace at situations. If we were to be judged in accordance to the regulation, very few does not conflict through a lot in justice. with the help of Christ, you should wish for mercy previous what the regulation ought to require. Having lived a existence right here on earth Jesus is established with what this is opt for to be powerless. including his blood he has offered the right to forgive sins. In Jesus Christ mercy has triumphed over justice!

Answer 4

Even though, i can do it all, i was wondering what part you are going to do yourself besides copying it down.
Your domain is: all real number because the denominator will never be equal to zero whatever the value of x