Calculus Question: Sketching a Function by Hand

Click For Summary

Discussion Overview

The discussion revolves around sketching the function $$f(x)=(3x-1)/(x-1)^2$$ by hand, focusing on finding intercepts, asymptotes, intervals of increase and decrease, local extrema, concavity, and inflection points. The scope includes calculus concepts such as limits, derivatives, and graphical analysis.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Participants calculate the x-intercept at $$f(0)=-1$$ and the y-intercept at $$f(1/3)=0$$.
  • There is a discussion on finding the horizontal asymptote, with some participants agreeing that it is at $$y=0$$ as $$x$$ approaches both $$\infty$$ and $$-\infty$$.
  • Vertical asymptote is identified at $$x=1$$, with participants confirming this based on the denominator of the function.
  • Intervals of increase and decrease are determined, with critical points found at $$x=-1/3$$, leading to a local minimum at $$f(-1/3)=-9/8$$.
  • Concavity is analyzed through the second derivative, with initial claims of intervals being concave up or down, which later prompted corrections regarding the signs due to the denominator being raised to the fourth power.
  • Participants discuss the existence of an inflection point at $$(-1,-1)$$, with some uncertainty about its classification due to the vertical asymptote at $$x=1$$.
  • Corrections and clarifications are made regarding the concavity intervals and the nature of the inflection point, with some participants revising their earlier calculations.

Areas of Agreement / Disagreement

While there is general agreement on the existence of the asymptotes and intercepts, there is some disagreement and uncertainty regarding the intervals of concavity and the classification of the inflection point at $$(-1,-1)$$. The discussion remains unresolved on the final classification of this inflection point due to differing interpretations of concavity changes.

Contextual Notes

Participants note limitations in their calculations, particularly regarding the signs in the concavity intervals and the implications of the vertical asymptote on the inflection point. There are unresolved mathematical steps related to the classification of the inflection point.

Pull and Twist
Messages
48
Reaction score
0
Hey all, I'm currently working a Calculus problem that requires me to sketch $$f(x)=(3x-1)/(x-1)^2$$ without using a calculator.

Part A: Requires that we find;

i. x-intercept:
$$f(0)=(3*0-1)/(0-1)^2=-1$$

So $$f(0)=-1$$

ii. y-intercept:
$$f(x)=0$$ therefore; $$3x-1=0$$ or $$x=1/3$$

So $$f(1/3)=0$$

iii. horizontal asymptote:
This is where I get stuck...

My initial thought is to use limits of x as it approaches $$\infty$$.

$$\lim_{{x}\to{\infty}}f(x)=(3x-1)/(x-1)^2$$

$$\lim_{{x}\to{\infty}}f(x)=(3x-1)/(x^2-2x+1)$$

$$\lim_{{x}\to{\infty}}f(x)=(3/x-1/x^2)/(1-2/x+1/x^2)$$

$$f(x)=(3/\infty-1/\infty^2)/(1-2/\infty+1/\infty^2)$$

$$f(x)=(0-0)/(1-0+0)=0$$

So the horizontal asymptote is at y=0 when we approach $$\infty$$??

Also, if I then take $$\lim_{{x}\to{-\infty}}$$ of the function I would also get 0??

iii. vertical asymptote:

$$f(x)=(3x-1)/(x-1)^2$$

$$(x-1)^2=0$$

$$x=1$$

So then the vertical asymptote is at $$x=1$$?

Part B: Find the intervals of increase and decrease, and the local and maximum and minimum values.

$$f(x)=(3x-1)/(x-1)^2$$

$$f'(x)=((x-1)^2(3)-(3x-1)(2(x-1)*(1)))/((x-1)^2)^2$$

$$f'(x)=((3(x-1)^2)-(2(3x-1)(x-1)))/(x-1)^3$$

$$f'(x)=(3x-3-6x+2)/(x-1)^3$$

$$f'(x)=(-3x-1)/(x-1)^3$$

So then we can solve for the Critical Points:

We know $$x\ne1 $$and $$-3x-1=0 \implies x=-1/3$$

If we then take $$f(-1/3)=(3x-1)/(x-1)^2$$ we get $$f(-1/3)=-9/8$$

So our intervals are $$(-\infty,-1/3),(-1/3,1),(1,\infty)$$

[table="width: 500, class: grid"]
[tr]
[td][/td]
[td]$$-3x-1$$[/td]
[td]$$x-1$$[/td]
[td]$$f'(x)$$[/td]
[/tr]
[tr]
[td]$$(-\infty,-1/3)$$[/td]
[td]$$+$$[/td]
[td]$$-$$[/td]
[td]decreasing[/td]
[/tr]
[tr]
[td]$$(-1/3,1)$$[/td]
[td]$$-$$[/td]
[td]$$-$$[/td]
[td]increasing[/td]
[/tr]
[tr]
[td]$$(1,\infty)$$[/td]
[td]$$-$$[/td]
[td]$$+$$[/td]
[td]decreasing[/td]
[/tr]
[/table]

Based on the table, we can tell which intervals are increasing and decreasing and that the local minimum is $$f(-1/3)=-9/8$$ and since we have a vertical asymptote at 1, there is no local max.

Part C: Find the intervals of concavity, and the inflection points of $$f$$.

$$f'(x)=(-3x-1)/(x-1)^3$$

$$f''(x)=(((x-1)^3)(-3)-(-3x-1)(3(x-1)^2*(1)))/((x-1)^3)^2$$

$$f''(x)=((-3(x-1)^3)-3(x-1)^2(-3x-1)/(x-1)^6$$

$$f''(x)=(-3(x-1)^2((x-1)+(-3x-1)))/(x-1)^6$$

$$f''(x)=(-3(-2x-2))/(x-1)^4$$

$$f''(x)=(6(x+1))/(x-1)^4$$

So then we can solve for the Critical Points:

We know $$x\ne1 $$and $$x+1=0 \implies x=-1$$

If we then take $$f(-1)=(3x-1)/(x-1)^2$$ we get $$f(-1)=-1$$

So our intervals are $$(-\infty,-1),(-1,1),(1,\infty)$$

[table="width: 500, class: grid"]
[tr]
[td][/td]
[td]$$x+1$$[/td]
[td]$$x-1$$[/td]
[td]$$f''(x)$$[/td]
[/tr]
[tr]
[td]$$(-\infty,-1)$$[/td]
[td]$$-$$[/td]
[td]$$-$$[/td]
[td]concave up[/td]
[/tr]
[tr]
[td]$$(-1,1)$$[/td]
[td]$$+$$[/td]
[td]$$-$$[/td]
[td]concave down[/td]
[/tr]
[tr]
[td]$$(1,\infty)$$[/td]
[td]$$+$$[/td]
[td]$$+$$[/td]
[td]concave up[/td]
[/tr]
[/table]

Based on the table, we can tell the concavity of the intervals and that an inflection point exists at $$f(-1)=-1$$ and since we have a vertical asymptote at 1, there is no inflection at 1.

Part D: Combine all the information from Part A-C and sketch the function.

I am currently working on this part and will post it once I am done.

Will someone let me know if I approaching this correctly?? Thank you.
 
Last edited:
Physics news on Phys.org
Hi PullandTwist,

iii. horizontal asymptote:
This is where I get stuck...

My initial thought is to use limits of x as it approaches $$\infty$$.

$$\lim_{{x}\to{\infty}}f(x)=(3x-1)/(x-1)^2$$

$$\lim_{{x}\to{\infty}}f(x)=(3x-1)/(x^2-2x+1)$$

$$\lim_{{x}\to{\infty}}f(x)=(3/x-1/x^2)/(1-2/x+1/x^2)$$

$$f(x)=(3/\infty-1/\infty^2)/(1-2/\infty+1/\infty^2)$$

$$f(x)=(0-0)/(1-0+0)=0$$

So the horizontal asymptote is at y=0 when we approach $$\infty$$??

You're doing a great job here in finding for the horizontal asymptote for the given rational function and yes, the horizontal asymptote for this problem is at $y=0$. (Yes)
iii. vertical asymptote:

$$f(x)=(3x-1)/(x-1)^2$$

$$(x-1)^2=0$$

$$x=1$$

So then the vertical asymptote is at $$x=1$$?

Correct. Vertical asymptotes are vertical lines which correspond to the zeros of the denominator of a rational function and in our case, $x=1$ is the zero of $(x-1)^2$.

PullandTwist

Part B: Find the intervals of increase and decrease, and the local and maximum and minimum values.[TABLE="class: grid, width: 500"]
[TR]
[TD][/TD]
[TD]$$-3x-1$$[/TD]
[TD]$$x-1$$[/TD]
[TD]$$f'(x)$$[/TD]
[/TR]
[TR]
[TD]$$(-\infty,-1/3)$$[/TD]
[TD]$$+$$[/TD]
[TD]$$-$$[/TD]
[TD]decreasing[/TD]
[/TR]
[TR]
[TD]$$(-1/3,1)$$[/TD]
[TD]$$-$$[/TD]
[TD]$$-$$[/TD]
[TD]increasing[/TD]
[/TR]
[TR]
[TD]$$(1,\infty)$$[/TD]
[TD]$$-$$[/TD]
[TD]$$+$$[/TD]
[TD]decreasing[/TD]
[/TR]
[/TABLE]

Based on the table, we can tell which intervals are increasing and decreasing and that the local minimum is $$f(-1/3)=-9/8$$ and since we have a vertical asymptote at 1, there is no local max.

Your answers are all correct for this part as well.

Part C: Find the intervals of concavity, and the inflection points of $$f$$.[TABLE="class: grid, width: 500"]
[TR]
[TD][/TD]
[TD]$$x+1$$[/TD]
[TD]$$x-1$$[/TD]
[TD]$$f''(x)$$[/TD]
[/TR]
[TR]
[TD]$$(-\infty,-1)$$[/TD]
[TD]$$-$$[/TD]
[TD]$$-$$[/TD]
[TD]concave up[/TD]
[/TR]
[TR]
[TD]$$(-1,\color{red}{-1}\color{black})$$[/TD]
[TD]$$+$$[/TD]
[TD]$$-$$[/TD]
[TD]concave down[/TD]
[/TR]
[TR]
[TD]$$(1,\infty)$$[/TD]
[TD]$$+$$[/TD]
[TD]$$+$$[/TD]
[TD]concave up[/TD]
[/TR]
[/TABLE]

Based on the table, we can tell the concavity of the intervals and that an inflection point exists at $$f(-1)=-1$$ and since we have a vertical asymptote at 1, there is no inflection at 1.

Note that you missed out the negative sign on the y-value for the inflection point $(-1,\,-1)$.

I say $(-1,-1)$ is the inflection point since an inflection point is where a curve changes from concave upward to concave downward or vice verse, which you have already proved so.
 
I just realized I calculated my concavity incorrectly. This corrected table should reflect the proper results. I forgot that the denominator was being raised to the power of 4, thus making any negative x value positive.

[table="width: 500, class: grid"]
[tr]
[td][/td]
[td]$$x+1$$[/td]
[td]$$(x-1)^4$$[/td]
[td]$$f''(x)$$[/td]
[/tr]
[tr]
[td]$$(-\infty,-1)$$[/td]
[td]$$-$$[/td]
[td]$$+$$[/td]
[td]concave down[/td]
[/tr]
[tr]
[td]$$(-1,1)$$[/td]
[td]$$+$$[/td]
[td]$$+$$[/td]
[td]concave up[/td]
[/tr]
[tr]
[td]$$(1,\infty)$$[/td]
[td]$$+$$[/td]
[td]$$+$$[/td]
[td]concave up[/td]
[/tr]
[/table]
 
anemone said:
Hi PullandTwist,

Note that you missed out the negative sign on the y-value for the inflection point $(-1,\,-1)$.

I say $(-1,-1)$ is the inflection point since an inflection point is where a curve changes from concave upward to concave downward or vice verse, which you have already proved so.

That was actually meant to signify the interval I was checking for concavity, so it should be (-1,1) based on the critical points I used. I determined the inflection based on where the signs changed in the table, which gave me x=-1 and x=1, but since I already new that we have a vertical asymptote at 1, the inflection at 1 could be ignored.

See image below...

View attachment 3527
 

Attachments

  • concavity.jpg
    concavity.jpg
    24.4 KB · Views: 120
PullandTwist said:
I just realized I calculated my concavity incorrectly. This corrected table should reflect the proper results. I forgot that the denominator was being raised to the power of 4, thus making any negative x value positive.

[TABLE="class: grid, width: 500"]
[TR]
[TD][/TD]
[TD]$$x+1$$[/TD]
[TD]$$(x-1)^4$$[/TD]
[TD]$$f''(x)$$[/TD]
[/TR]
[TR]
[TD]$$(-\infty,-1)$$[/TD]
[TD]$$-$$[/TD]
[TD]$$+$$[/TD]
[TD]concave down[/TD]
[/TR]
[TR]
[TD]$$(-1,1)$$[/TD]
[TD]$$+$$[/TD]
[TD]$$+$$[/TD]
[TD]concave up[/TD]
[/TR]
[TR]
[TD]$$(1,\infty)$$[/TD]
[TD]$$+$$[/TD]
[TD]$$+$$[/TD]
[TD]concave up[/TD]
[/TR]
[/TABLE]

Ah, sorry about not checking out properly the signs that you got in the intervals that you set in. I worked this problem out on my paper and I noticed that $(-1,\,-1)$ is a valid inflection point.

Okay, by setting the second derivative equals 0 gives us $x=-1$ and substituting this back to the original rational function to get the corresponding $f(x)$ value, we see that this yields $f(-1)=-1$. So, $(-1,\,-1)$ may or may not be an inflection point.

Since inflection points are where the function changes concavity and since we have gotten only one critical point with its $x$-coordinate as $-1$, we only have two intervals to check: $(-\infty,-1]$ and $[-1,\infty)$.

Now, according to the table with the corrected result, it shows that the curve does change from concave downward to concave upward at $(-1,\,-1)$, hence, $(-1,\,-1)$ is an inflection point.
 
After all that, this is the graph I was able to get...

Oh, and I also reversed my x & y intercepts in my original post. Ooops... (Mmm)

View attachment 3528
 

Attachments

  • finalgraphsmall.jpg
    finalgraphsmall.jpg
    54.2 KB · Views: 136

Similar threads

High School Understanding exponentiation and logarithms together
  • · Replies 12 ·
Replies
12
Views
2K
Undergrad Showing That a Function Does Not Have Two Distinct Roots
  • · Replies 3 ·
Replies
3
Views
2K
MHB What is the Slant Asymptote of $\dfrac{4x^3-10x^2-11x-1}{x^2-3x}$?
  • · Replies 3 ·
Replies
3
Views
2K
Graduate Solving a power series
  • · Replies 3 ·
Replies
3
Views
4K
Graduate Summing over continuum and uncountable numerocities
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad ##f(2x)=f^2(x)-2f(x)-1/2## then find ##f(3)##
  • · Replies 17 ·
Replies
17
Views
3K
High School Can We Simplify the Taylor Series Expansion for e^(f(x,y))?
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Express the limit as a definite integral
  • · Replies 16 ·
Replies
16
Views
4K
High School Why is this definite integral a single number?
  • · Replies 20 ·
Replies
20
Views
4K
Graduate Multiplying divergent integrals using Hardy fields approach
  • · Replies 3 ·
Replies
3
Views
2K