Derivative Homework Question?

Question

Hi there,
I've been having a little bit of trouble with this derivative homework of mine.
It says that f(x) = (-4x -3) / (x^2 -x +4)
I need to find the zeros/roots of this function, and I only got a zero when x= - 3/4...are there any others?

Also, we need to find the intervals on which the function is increasing/decreasing.
I found the first derivative to be (4x^2 +6x -19) / (x^2 -x +4)^2
But I cannot find any x-value where f ' (x) is 0 or undefined...
Therefore I can't find when the function is increasing or decreasing.

I had no trouble with the second derivative and inflection points, but another question asks if the original function has any vertical asymptotes. I cannot find any, but did anyone else?

Answer 1

You have the right zero and the right derivative for f(x).

To find the intervals where the function is increasing or decreasing, you need to know more about the denominator as the range of f(x) is not all real numbers x. Namely, f(x) is not defined where x^2 - x + 4 = 0.

As f'(x) = (4x^2 + 6x - 19)/(x^2 - x + 4)^2

f'(x) = 0 when 4x^2 + 6x - 19 = 0

Solving this gives

x = [+/-sqrt(85) - 3]/4

Since this doesn't also solve the denominator

x^2 - x + 4 = 0

Then f'(x) has 2 well defined zeros (critical points).

So let's say you have the 2 critical points, x1 and x2.

You use the 2nd derivative to find whether it is a maximum, minimum or inflection point.

If x1 is a maximum then the interval to the left (xx1) must be decreasing and vice versa for a minimum.

When you look for vertical asymptotes you think in terms of derivatives which go to infinity at the limit so look for places where f'(x) -> infinity. The first place you should look is where you have division by zero.

Since f(x) has a polynomial denominator, you will have at least one vertical asymptote.

Since the denominator of f'(x) is (x^2 - x + 4)^2 you vertical asymptotes are solutions of

(x^2 - x + 4)^2 = 0

x^2 - x + 4 = 0

which has no real solutions so you have no real vertical asymptotes.

Answer 2

yes... there is only one zero... x = -3/4


to find the critical numbers... you need to solve for the zero of the quadratic: 4x² + 6x - 19

by quadratic formula:
x = [-6 ± √(36 + 304)]/8
= -3.05 & 1.55

to determine on which interval it is increasing or decreasing...
use test points...

finally... there will be no vertical asymptote because the original function has no zero in the denominator... you can check that by getting the discriminant of the denominator...


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Answer 3

Your first derivative is correct. Use the quadratic formula to find where the numerator is 0.

The zeros of the denominator are not real, so there are no vertical asymptotes.

Answer 4

fx=3x^2-4y^3[10xy]-5x^2y[6x]/[(3x^2)-4y^... =30x^3y-40xy^4-30x^3y/[(3x^2)-4y^3)] =-40xy^4/[(3x^2-4y^3)]^2 fy=3x^2-4y^3[5x^2]-5x^2y[-12y^2]/[(3x^... 15x^4-20x^2y^3+60x^2y^3/[(3x^2-4y^3)]^... [15x^4+40x^2y^3]/[(3x^2-4y^3)]^2