Question

Please Help!!

Find the critical points and the intervals on which the function is increasing or decreasing. Use the First Derivative Test to determine whether each critical point is a local min or a local max or neither.

Equation posted below:

Answer 1

\[y=\frac{x^{3}}{ x^2-3 }\]

Answer 2

1. What do you know about the definition of "critical value"? How do you obtain the critical values of a given function?
2. Once you have the critical value or values, how do you obtain the y-value corresponding to each? How do you write "critical points," once you have the critical values and the corresponding y-values?

Answer 3

What is the difference between critical values and critical points? and how are they related? I know you need to take the first derivative to get one. Do you have to use the quotient rule to find the first derivative? I need a walk through, this problem is confusing me.

Answer 4

Hi, Nate! A "critical value" is a root the equation that results when we set the derivative of a function = to 0. You'll probably remember that finding the critical values of a function leads to specifying where possible maxima/minima/horizontal points of inflection of the curve are located, in terms of their x-values. For example: if f(x) = x^2 + 2x, f'(x) = 2x + 2.
Setting this equal to zero, we get 2x + 2 = 0, or 2(x+1) = 0, or x = -1. That's a "critical value" and is the only one in this example of mine.
A "critical point" has two coordinates, like any other point in the plane: (x, f(x)). Here, x represents one critical value, and f(x) represents the value of the function at that x.

Notice that the given function, y=[x^2)]/(x^2-3), is a quotient. The quickest way to find the derivative of this quotient is to apply the quotient rule. Please look that up and write it down if you don't feel completely comfortable with this rule.

Please do the best you can to find the derivative of y=[x^2)]/(x^2-3) using the Quotient Rule. Then I could give you more meaningful feedback on your work.

I've done that myself, obtaining the derivative f'(x) = -6x/[x^2-3)^2.

Set this equal to zero and solve for the critical value x (there's only one in this case).

Now evaluate the given function at this critical value (x value), and write the result as a point: (x, f(x)). That's your (sole) critical point.

Answer 5

Although I can't promise fast feedback, I'll certainly answer any notes you send me as soon as possible.

Answer 6

working on it now...

Answer 7

Good. Let me summarize the quotient rule: if u and v are separate functions of x, then the derivative of the quotient y = u/v is y' = v*u' - u*v
--------
v^2

Answer 8

Sorry about the lousy formatting. That result should be y' = (v*u' - u*v)/(v^2).

Answer 9

In the problem you've presented, the numerator, u, is u = x^2; the denominator, v, is v = (x^2 - 3). Find u' and v'. Then plug u, v, u' and v' into the given formula for the derivative of a quotient u/v.

Answer 10

Nate? Hey, are you familiar with the process of determining the intervals on which a given function is increasing or decreasing? Hint: this info comes from the first derivative.

Answer 11

Hey, so I got a completely different first derivative:

I got \[\frac{ x^2(x^2-9) }{ (x^2-3)^2 }\] after using the quotient rule. how did you get -6x on top? I know how to do derivatives and I think that is wrong. Anyway, I set my derivative to 0 and solved for x getting 3 or -3. Now what?

Answer 12

\[\frac{x^{2}}{x^{2}-3}\] quotient rule
\[\frac{ 2x(x^{2}-3) - x^{2}(2x) }{ (x^{2}-3)^{2} } \implies \frac{ 2x^{3} - 6x - 2x^{3} }{ (x^{2}-3)^{2} } = \frac{ -6x }{ (x^{2}-3)^{2} }\]

Answer 13

well its an interesting curve

vertical asymptotes at

\[x = \pm \sqrt{3}\]

and a horizontal asymptote at y = 1

so to find the critical values set f'(x) = 0 and solve for x

but really all you are solving is 0 = -6x

since if the numerator is zero the fraction is zero...

all you need to do is test either side of the stationary point... and look at what the value of f'(x) is....

hope this helps

Answer 14

Nice work, Concentration and Campbell! Thanks for your contributions.

Nate: Now that you have ONE critical value, x = 0, draw a number line and graph x = 0 on that line (just place a mark where x = 0).

This ONE critical value divides the number line in to TWO intervals: x<0 and x>0.
Please choose a representative number from each interval (such as -1 from x<0) and determine the sign of the first derivative. Concentration has supplied us with the correct formula for that derivative; simply substitute -1 for x and say whether the result is + or -.
Then do the same thing for the other interval.

Rule: If the sign of the derivative is negative on an interval, the function itself is decreasing on that interval. If positive, the function is increasing.

Rule: If the function is decr on the left and incr on the right of a critical point, we have a local minimum there. If incr on the left and decr on the right, we have a local maximum there.

Do come up with more questions if need be. Good luck, and thanks for your perseverance!

Answer 15

Campbell is right on target when he identifies the vertical asymptotes as being plus or minus Sqrt(3). These two x-values, along with x = 0, constitute the "critical values" of this function. Sorry I didn't think of that earlier.
With three critical values, we'll have four intervals on the number line. As before, choose a representative x value from each interval and determine the sign of the first derivative for each such value.

Answer 16

@concentrationalizing Yeah that quotient rule is backwards, you took the derivative of g(x) (x^2-3) second, you take that first. you did f(x) * g '(x) - g(x) * f '(x) over g(x)^2 the quotient rule is g(x) * f '(x)- f(x) * g(x) over g(x)^2

That won't give you the correct derivative, I believe the derivative I got earlier is the correct one.

Answer 17

Nate: Please go back and check out what "Concentrationalizing" has done. I agree with his result. Have you a teacher, friend, fellow student or mentor nearby with whom you could check out why you're getting a result different from that which "Concentrationalizing" and I have both obtained?

Have you checked your personal formula for the derivative of a quotient with one in your textbook or notes?

Answer 18

Yes, there's a typo in the formula for the derivative of a quotient that I typed 17 hours ago: y' = (v*u' - u*v)/(v^2). The second term in the numerator should be u*v' (not u*v). Thank you for pointing this out. However, I do stand by my application of the quotient rule to the problem we're discussing.