Question

For the following function, f, determine the domain, locate the critical points of f; use the First Derivative Test to locate local extrema; identify any absolute extrema; determine the intervals for which f is increasing/decreasing; determine the intervals for which f is concave up/down; identify and inflection points; determine any aspymptotes; and sketch the function.

f(x) = x
------
x^2 - 9

Answer 1

\[f(x) = \frac{ x }{ x ^{2} - 9 }\]

Answer 2

What's your plan? The Domain is an eyeball problem. You should just look at it and shout it out.

ALL REAL NUMBERS except x = 3 and x = -3. Tell me where I got that so easily.

Answer 3

yah, I got that - you just set the denominator equal to zero. I have the derivative but need help with critical points. 3 and -3 are not critical points and when I set the numerator of the derivative to zero i get an imaginary number.

Answer 4

Okay, since Critical Points for a rational function are REAL numbers, what does that say about the critical points as exposed by the 1st Derivative?

Answer 5

that they dont exist? ... this is where im confused

Answer 6

There is nowhere on f(x), for REAL x, where the 1st derivative is zero. Perfect.

Let's move on to the 2nd Derivative.

Answer 7

The second derivative is used to find the inflection points and use them to find where concavity changes.

Answer 8

Excellent. Let's see it.

Answer 9

okay, so \[f \prime(x) = \frac{ -x ^{2} - 9 }{ (x ^{2} -9)^{2} }\]

then \[f \prime \prime(x) = \frac{ (x ^{2}-9)(-2x) - (-x ^{2} -9)(4x(x ^{2}-9)) }{ (x ^{2} -9)^{4}}\]
right?

Answer 10

So close! Make that first \((x^{2} - 9)\) squared.

Answer 11

oops! It is on my paper.. . just didnt type it that way :)

Answer 12

would inflection points be -3,3,0?

Answer 13

Very good. I get \(f"(x) \ \dfrac{2x(x^{2}+ 27)}{(x^{2} - 9)^{3}}\)

Tell me why x = 3 and x = -3 can be inflection points. They are not in the Domain!

Answer 14

because they are points where change if it is decreasing or increasing even though it may not be differentiable

Answer 15

Very good answer. If it's positive on one side and negative on the other, something must have happened.

Answer 16

and then I just need to plug in numbers to the second derivative on the intervals (below) to find of they are positive or negative, which will give me the concavity?

\[(-\infty, -3)(-3, 0) (0, 3) (3, \infty) \]

Answer 17

So we have for Concavity

\(Downward < -3 < Upward < 0 < Downward < 3 < Upward\)

Okay, we already have the x- and y-intercepts (0,0). We have only an horizontal asymptote and we're ready to draw.

Answer 18

perfect :) Thank you so much! This all seemed so easy when it was separate and easy example problems in class and in my notes!

Answer 19

One piece at a time. Just wade through it systematically. Very, very good work.

Answer 20

thanks again :)