# Thread: Application of Limits and Derivatives to Rational Functions

1. ## Application of Limits and Derivatives to Rational Functions

Hey guys, please tell me if I'm correct in my calculations
Given the function $\frac{x^2}{x-1}$

a) Determine the domain and all the intercepts of the function. Does the function have any symmetry?

This is what I did: Domain is all real #s except 1, because x-1=0, x=1.

intercepts are/is: (0,0) because $\frac{(0)^2}{(0)-1}$=0

it's symmetric because $\frac{(-x)^2}{(-x)-1}$= $\frac{x^2}{-x-1}$

Is this right?

b) Determine $\lim_{x \to \infty}$ f(x) and $\lim_{x \to -\infty}$ f(x). Determine all horizontal and vertical asymptotes of the function.

I got that lim x-->inf it goes to inf and lim x-->-inf goes to -inf.
Vertical asymptote is 1 as shown above.

Can someone show me how you get the horizontal asymptote for this please?

c) find all critical numbers.

Quotient rule:
$\frac{(x-1)(2x)-(x^2)(1)}{(x-1)^2} = \frac{x^2-2x}{(x-1)^2 }$

$x^2-2x=0, x=0,2$ critical numbers are 0 and 2

d) Determine the intervals on which the function is increasing and where it is decreasing. Find all maximum and minimum values of f.

I use a chart.
______0__2
x __|- | + | +
x-2 |- | - | +
f '_ |+ | - | +
___up, down, up. so there must be a maximum at 0 and a minimum and 2. right?

2. Originally Posted by RezMan
Hey guys, please tell me if I'm correct in my calculations
Given the function $\frac{x^2}{x-1}$

a) Determine the domain and all the intercepts of the function. Does the function have any symmetry?

This is what I did: Domain is all real #s except 1, because x-1=0, x=1.

intercepts are/is: (0,0) because $\frac{(0)^2}{(0)-1}$=0

it's symmetric because $\frac{(-x)^2}{(-x)-1}$= $\frac{x^2}{-x-1}$

Is this right?
A function is even if f(-x) = f(x) and odd if f(-x) = -f(x). In your case
$f(-x) = \frac{x^2}{-x - 1}$

Is this even, odd, or neither?

Originally Posted by RezMan

b) Determine $\lim_{x \to \infty}$ f(x) and $\lim_{x \to -\infty}$ f(x). Determine all horizontal and vertical asymptotes of the function.

I got that lim x-->inf it goes to inf and lim x-->-inf goes to -inf.
Vertical asymptote is 1 as shown above.

Can someone show me how you get the horizontal asymptote for this please?
[COLOR=Red][COLOR=Black]
You get a horizontal asymptote when $\lim_{x \to \pm \infty}f(x) \to \text{constant}$. If this case your limits are either positive or negative infinity. So you have no horizontal asymptotes.

-Dan

Edit: You do have a "slant" asymptote if you need to look for one of those. The slant asymptote is y = x + 1.

3. Originally Posted by RezMan
c) find all critical numbers.

Quotient rule: [/COLOR][/COLOR] $\frac{(x-1)(2x)-(x^2)(1)}{(x-1)^2} = \frac{x^2-2x}{(x-1)^2 }$

$x^2-2x=0, x=0,2$ critical numbers are 0 and 2
Don't forget to set your denominator to 0. x = 1 is also a critical number.

Originally Posted by RezMan
d) Determine the intervals on which the function is increasing and where it is decreasing. Find all maximum and minimum values of f.

I use a chart.
______0__2
x __|- | + | +
x-2 |- | - | +
f '_ |+ | - | +
___up, down, up. so there must be a maximum at 0 and a minimum and 2. right?
Redo your chart with the x = 1 in there.

-Dan

4. Thank you so much Dan!
Originally Posted by topsquark
Is this even, odd, or neither?
I guess it's neither, right? I just thought there was some rule where you only pay attention to the top.

Originally Posted by topsquark
You do have a "slant" asymptote if you need to look for one of those. The slant asymptote is y = x + 1.

5. Originally Posted by RezMan
Thank you so much Dan! I guess it's neither, right? I just thought there was some rule where you only pay attention to the top.
That's correct, it is neither.

Originally Posted by RezMan
A slant asymptote, otherwise known as an "oblique" asymptote, is similar to a horizontal asymptote, except that it is off at an angle. (I'll leave the pictures to the link.)

To sum up, you can tell if a function has a slant asymptote by looking at the degree of the numerator (dn) and the degree of the denominator (dd). If dn - dd = 1 then there is a slant asymptote. We get the form of the asymptote by division.

For example, in your function you have
$f(x) = \frac{x^2}{x - 1}$

The degree of the numerator is 2 and the denominator 1. 2 - 1 = 1 so you have a slant asymptote. Unfortunately I can't code the long division, so I'll have to leave that to you. Here's the result.
$f(x) = \frac{x^2}{x - 1} = x + 1 + \frac{1}{x - 1}$

As with the test for a horizontal asymptote we look at the form as x goes to infinity. The 1/(x - 1) term becomes inconsequential, so the result is y = x + 1. That's your slant asymptote.

-Dan

Edit: You can actually do the division without using long division. Notice that
$\frac{x^2}{x - 1} = \frac{x^2 - 1 + 1}{x - 1} = \frac{x^2 - 1}{x - 1} + \frac{1}{x - 1} = (x + 1) + \frac{1}{x - 1}$

6. Originally Posted by topsquark
Redo your chart with the x = 1 in there.
______0__1__2
x __|- | + | + |+
x-2 |- | - | - | +
f '_ |+ | - | - | +
__up, down, down, up. so there's local maximum at 0 and local minimum at 2, and no absolute max/min?

I can't thank you enough, this exercise is so difficult for me. There's more

e) Determine where f is concave up and concave down.

I figured that f''' = $\frac{2}{(x-1)^3}$. so when plugging in 0 f'' is concave down, and concave up at 2.

7. Originally Posted by RezMan
______0__1__2
x __|- | + | + |+
x-2 |- | - | - | +
f '_ |+ | - | - | +
__up, down, down, up. so there's local maximum at 0 and local minimum at 2, and no absolute max/min?

I can't thank you enough, this exercise is so difficult for me. There's more

e) Determine where f is concave up and concave down.

I figured that f''' = $\frac{2}{(x-1)^3}$. so when plugging in 0 f'' is concave down, and concave up at 2.
The chart looks good now. A point: A function is concave up or down on an interval. So f(x) is concave up whenever f''(x) is negative and concave down when f''(x) is positive.

You can either figure that out by punching numbers into your second derivative (or sketch it, the second derivative is easy enough to sketch) or by using your chart. You know where the turning points are and you know what the slope is doing. So, for example, the slope is increasing on (-infinity, 0), it has a local maximum at x = 0 and then decreases on (0, 1). What must the concavity of the function be on (-infinity , 1)? Think about this for the second branch of the function, (1, infinity).

-Dan

8. is the minima/maxima thing I said correct though? That there's a local min&max but no absolute?

9. Originally Posted by RezMan
is the minima/maxima thing I said correct though? That there's a local min&max but no absolute?
When f''(x) is positive, the function is concave down. When f''(x) is negative, the function is concave up. It is the reverse of what you might expect. So you have your concavities reversed.

-Dan

10. Originally Posted by topsquark
When f''(x) is positive, the function is concave down. When f''(x) is negative, the function is concave up. It is the reverse of what you might expect. So you have your concavities reversed.
-Dan
Are you sure? Because the graph makes sense and also this website says the opposite
Concavity of Graphs

11. Originally Posted by RezMan
Are you sure? Because the graph makes sense and also this website says the opposite
Concavity of Graphs
(sighs) I think what I was reaching for is that we have a local minimum on a concave up domain and a local maximum on a concave down domain. Sorry for the poor information!

-Dan

12. Don't hit yourself against the wall dan!

So are you saying that my min/max is wrong?