How to determine the intervals of decrease or increase

In summary, the conversation discusses a problem with curve sketching and finding various characteristics of a given curve. These include the domain, intercepts, asymptotes, intervals of increase or decrease, concavity, and points of inflection. However, there are some errors in the solutions provided, such as incorrect calculations and not following the proper steps for finding the intercepts and derivatives. The conversation also clarifies the function being used and addresses the need to find a horizontal asymptote, even if it is a slant one.
  • #1
F.B
83
0
Im having a problem with curve sketching i can't figure out how to do this problem. I am confused.

Heres the question

1. For the following curve find:
a)the domain
b)the intercepts
c)the asymptotes
d)intervals of increase or decrease: local maximum and minimum
e)concavity and the point of inflection

x^2 - x - 1 / x-1


Heres my work.

a) the domain is x = 1
b)Let y=0
y = 0 - 0 - 1/ 0-1
y=1
Let x = 0
x^2 - x - 1 = 0
x = 1.62 or x = 0.62

c) Vertical asymptote:
x = 1
Horizontal asymptote
x^2 - x - 1 / x-1
y = x - 1/ x-1
Therefore it has an slant asympote at y = x

Heres where it gets fun

d) y = x^2 - x - 1/x-1
y'= x^2 - 2x + 2/ (x-1)^2
let y'= 0
x^2 - 2x - 2 = 0
x = 1 +/- i

Do i say that it doesn't have an maximum or minimum point because they are
imaginary?

e) y'' = -2/(x-1)^3
Let y''=0
-2 cannot equal 0

So there are no points of inflection, but then how do i determine the
concavity when my first derivative is based on imaginary numbers.

I have never encountered one like this. I don't know how to determine the intervals of decrease or increase because of i, so can you please help me.
 
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  • #2
F.B said:
Im having a problem with curve sketching i can't figure out how to do this problem. I am confused.

Heres the question

1. For the following curve find:
a)the domain
b)the intercepts
c)the asymptotes
d)intervals of increase or decrease: local maximum and minimum
e)concavity and the point of inflection

x^2 - x - 1 / x-1


Heres my work.

a) the domain is x = 1
How do you get that? The domain is the set of all x where
f(x) = x^2 - x - 1 / x-1

So x = 1 is NOT part of the domain of f(x).
b)Let y=0
You must mean let x = 0 since you are trying to find the y intercept. The y intercept is a point (0,y)
Let x = 0
Again, this should be y = 0 to find the x intercept.
x^2 - x - 1 = 0
x = 1.62 or x = 0.62
These are the correct solutions to the quadratic equation.
c) Vertical asymptote:
x = 1
Correct
Horizontal asymptote
x^2 - x - 1 / x-1
y = x - 1/ x-1
Therefore it has an slant asympote at y = x
You will have to explain why you did this. I don't follow this.
Heres where it gets fun

d) y = x^2 - x - 1/x-1
y'= x^2 - 2x + 2/ (x-1)^2
let y'= 0
x^2 - 2x - 2 = 0
x = 1 +/- i
You lost me there.

AM
 
  • #3
F.B said:
Im having a problem with curve sketching i can't figure out how to do this problem. I am confused.

Heres the question

1. For the following curve find:
a)the domain
b)the intercepts
c)the asymptotes
d)intervals of increase or decrease: local maximum and minimum
e)concavity and the point of inflection

x^2 - x - 1 / x-1


Heres my work.

a) the domain is x = 1
Assuming that y= x^2- x- 1/(x-1), this is exactly backwards: the domain is all x except x= 1.

b)Let y=0
y = 0 - 0 - 1/ 0-1
y=1
As Andrew Mason pointed out, actually, you've let x= 0 to find the y-intercept.

Let x = 0
x^2 - x - 1 = 0
x = 1.62 or x = 0.62
No, you let y= 0 so you get the equation x^2- x- 1/(x-1)= 0 which is the same as x^2- x= 1/(x-1) or (x-1)(x^2-x)= (x-1)^2(x)= 0. The x intercepts are x= 0 and x= 1.

c) Vertical asymptote:
x = 1
Ok.

Horizontal asymptote
x^2 - x - 1 / x-1
y = x - 1/ x-1
Therefore it has an slant asympote at y = x
But the problem didn't say "slant asymptote"- it asked about a horizontal asymptote- and there is none. No, y= x is not a "slant asymptote" anyway. How did you go from y= x^2- x- 1/(x-1) to y= x- 1/(x-1)??
x^2- x is NOT equal to x!


Heres where it gets fun

d) y = x^2 - x - 1/x-1
y'= x^2 - 2x + 2/ (x-1)^2
?? Are you saying that the derivative of x^2 is x^2 and the derivative of -x is -2x? Surely you know better than that! Oh, and the derivative of -1/(x-1)= -(x-1)^(-1) is NOT 2/(x-1)^2, although I admit that's harder. For all of those you can use the fact that the derivative of
x^n is nx^(n-1).

let y'= 0
x^2 - 2x - 2 = 0
x = 1 +/- i

Do i say that it doesn't have an maximum or minimum point because they are
imaginary?
Go back and do the derivative correctly!

y'' = -2/(x-1)^3
Let y''=0
-2 cannot equal 0

So there are no points of inflection, but then how do i determine the
concavity when my first derivative is based on imaginary numbers.
Again, go back and do the first derivative correctly. You can't get the second derivative until you do that.

I have never encountered one like this. I don't know how to determine the intervals of decrease or increase because of i, so can you please help me.
For future reference, if you had a problem where the derivative was never 0, such as f(x)= x, then the derivative always has the same sign and so there is just one "interval of decrease or increase".
 
  • #4
To find the y-intercept you have to solve for y correct? Thats why you let x=0 to find the find what y is when equals 0. To find the x-intercept you have to let y be 0 and then you solve for x to determine what x is when y is 0. And Halls of Ivy maybe i should have made the question a bit clearer. The function is this:

y=(x^2-x-1)/(x-1), does this help?

For the horizontal asymptote we have to find one even it is a slant one.
By doing long division i get the following:

y = (x) - (1)/(x-1).

My derivatives are all right because when i use the question rule i get that same thing.

But i figured how to find the intervals of increase and decrease. Because the vertical asymptote x=1, i can use those as my intervals. So x<1 and x>1.

But i still have one problem. If i use those as my intervals, when i do that chart with the test values, do i sub a number <1 and >1 into the original function or the first derivative.

My other problem is determining the concavity. Because i don't have a point of inflection, how would i determine where the function is concave up and down.
 
  • #5
F.B said:
To find the y-intercept you have to solve for y correct? Thats why you let x=0 to find the find what y is when equals 0. To find the x-intercept you have to let y be 0 and then you solve for x to determine what x is when y is 0.

That is correct.

F.B said:
And Halls of Ivy maybe i should have made the question a bit clearer. The function is this:

y=(x^2-x-1)/(x-1), does this help?

Halls of Ivy's point still remains valid, the function is defined everywhere except at x = 1. Therefore, the domain of the function is [itex] x \in\Re \; , \; x \neq 1[/itex]

F.B. said:
My derivatives are all right because when i use the question rule i get that same thing.
I'm afraid you derivative is incorrect. Try and do it again, perhaps it would be easier to write it as;

[tex]f(x) = (x^{2} - x -1)(x-1)^{-1}[/tex]
 
Last edited:
  • #6
F.B said:
My other problem is determining the concavity. Because i don't have a point of inflection, how would i determine where the function is concave up and down.
The concavity is given by the sign of the second derivative. Watch out though: the fact that there are no zeroes doesn't mean the concavity is the same everywhere, remember your vertical asymptote!
 

1. What is the first step in determining the intervals of decrease or increase?

The first step is to graph the data points on a coordinate plane. This will help visualize the overall trend of the data.

2. How do you determine if there is a decrease or increase in the data?

To determine if there is a decrease or increase in the data, look for a consistent trend in the data points. If the data points are consistently moving upwards, there is an increase. If they are consistently moving downwards, there is a decrease.

3. How do you find the intervals of decrease or increase?

The intervals of decrease or increase can be found by identifying the points where the trend changes from increasing to decreasing or vice versa. These points are known as turning points and mark the beginning or end of an interval.

4. What is the significance of determining the intervals of decrease or increase?

Determining the intervals of decrease or increase allows us to understand the overall trend of the data and make predictions about future data points. It also helps identify any potential outliers or anomalies in the data.

5. Are there any mathematical formulas to determine the intervals of decrease or increase?

No, there are no specific formulas for determining the intervals of decrease or increase. It is a visual analysis of the data points and identifying the turning points. However, there are mathematical methods such as regression analysis that can help determine the overall trend of the data.

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