Finding Solutions to x Goes to Infinity and Beyond

[asdfsystema]

Homework Statement

Hey guys. I tried solving these on my own, but I cannot seem to find out how to do this.

23. What are the steps to finding

sqrt(3+6*x^2)/(2+5*x) as x goes to infinity?


41. Find the Horizontal Asymptotes for

17x/(x^4+1)^1/4 . I'm pretty sure its 17/1, but for some reason, my teacher asks for two, and I thought rational functions only have one horizontal asymptote...



62. f(x)= 4x^3+13x^2+11x+24 / x+3 when x<-3
f(x)= 3x^2+3x+A when -3 less than or equal to x

What is A in order for it to be continuous at -3?


79. f is continuous at (-inf, + inf)

f(y) = cy+3 range is (-inf,3)
f(y) = cy^2-3 range is (3,+inf)

what is C?


81. Let f(x) = {2x^2+3 x -65) / (x-5)

Show that f(x) has a removable discontinuity at x=5 and determine what value for f(5) would make f(x) continuous at x=5.
Must define f(5)=



Thank you so much in advance >< I've been doing these questions for 3 hours ... I need help to prepare for my calculus test next week !

 

[HallsofIvy]

Homework Statement

Hey guys. I tried solving these on my own, but I cannot seem to find out how to do this.

23. What are the steps to finding

sqrt(3+6*x^2)/(2+5*x) as x goes to infinity?

If you tried, what did you do? It is always better to show what you have tried. One thing you can do is say "if x is extremely large, we can ignore "3" in comparison with 6x² and "2" in comparison to 5x. What do you get if you just drop the 3 and 2?

A little more rigorous is to divide both numerator and denominator by x. While "going to infinity" is a nuisance, going to 0 is easy- and when x goes to infinity, 1/x goes to 0.

41. Find the Horizontal Asymptotes for

17x/(x^4+1)^1/4 . I'm pretty sure its 17/1, but for some reason, my teacher asks for two, and I thought rational functions only have one horizontal asymptote...

No, that is not necessarily so. "Horizontal asymptotes" occur is the graph approaches a horizontal line as x goes to infinity or negative infinity. What happens if x is a large negative number? For example, what is the value if x= -100000000?

62. f(x)= 4x^3+13x^2+11x+24 / x+3 when x<-3
f(x)= 3x^2+3x+A when -3 less than or equal to x

What is A in order for it to be continuous at -3?

What is the definition of continuous? In particular, what are the limits as x goes to -3 from above and below?

79. f is continuous at (-inf, + inf)

f(y) = cy+3 range is (-inf,3)
f(y) = cy^2-3 range is (3,+inf)

what is C?

This makes no sense. First you must mean "on" (-inf, +inf), not "at". Second, do you mean "domain" rather than "range"? And, finally, the only way that would make sense is if the two domains are (-inf, 3] and [3,+inf). Assuming those are correct, what are the limits as x goes to 3 from above and below?

81. Let f(x) = {2x^2+3 x -65) / (x-5)

Show that f(x) has a removable discontinuity at x=5 and determine what value for f(5) would make f(x) continuous at x=5.
Must define f(5)=

You must define f(5) to be the limit as x approaches 5. f does not have a value at x= 5, given by that formula, because the denominator is 0 at x= 5. But in order to have a limit at x= 5, the numerator must also be ____. And what does that tell you about factoring 2x²+ 3x- 65?

Thank you so much in advance >< I've been doing these questions for 3 hours ... I need help to prepare for my calculus test next week !

 

[asdfsystema]

HallsofIvy, thanks for replying. I will try to post up what I did here.

23. What are the steps to finding

sqrt(3+6*x^2)/(2+5*x) as x goes to infinity?

i figured out the answer is sqrt(2) x sqrt (3) / 5 I don't get what I'm supposed to do to the numerator with the square roots.

What I did is turn that function disregarding the +2 and +3 so I got sqrt 6x^2 / 5x ... and I'm not suer what I do next...

41. Find the Horizontal Asymptotes for

17x/(x^4+1)^1/4 . I'm pretty sure its 17/1, but for some reason, my teacher asks for two, and I thought rational functions only have one horizontal asymptote...[/quote]
No, that is not necessarily so. "Horizontal asymptotes" occur is the graph approaches a horizontal line as x goes to infinity or negative infinity. What happens if x is a large negative number? For example, what is the value if x= -100000000?

What I did first is fix the denominator and make it become 17x / x+1 because I multiply the roots. I disregard the +1 and get 17x/x ... This has two horizontal asymptotes?
I thought because the ratio of the coefficient of 17x and x is 1 , n=m=1 therefore it is 17/1 ? please correct me if I'm wrong . if x is -10000 then it is neg infinity .


62. f(x)= 4x^3+13x^2+11x+24 / x+3 when x<-3
f(x)= 3x^2+3x+A when -3 less than or equal to x

What is A in order for it to be continuous at -3?[/quote]
What is the definition of continuous? In particular, what are the limits as x goes to -3 from above and below?

continuous means there are no breaks , basically being able to draw it on a graph without lifting up pencil. also there is no discontinuity. as x goes to -3 when x<-3 it goes to 0?
I still don't understand what A has to be...


79. f is continuous at (-inf, + inf)

f(y) = cy+3 range is (-inf,3)
f(y) = cy^2-3 range is (3,+inf)

what is C?

This makes no sense. First you must mean "on" (-inf, +inf), not "at". Second, do you mean "domain" rather than "range"? And, finally, the only way that would make sense is if the two domains are (-inf, 3] and [3,+inf). Assuming those are correct, what are the limits as x goes to 3 from above and below?

Sorry I did mean on, and yes it is the domain. the limits as x goes to 3 from above is -inf ?




81. Let f(x) = {2x^2+3 x -65) / (x-5)

Show that f(x) has a removable discontinuity at x=5 and determine what value for f(5) would make f(x) continuous at x=5.
Must define f(5)=

You must define f(5) to be the limit as x approaches 5. f does not have a value at x= 5, given by that formula, because the denominator is 0 at x= 5. But in order to have a limit at x= 5, the numerator must also be ____. And what does that tell you about factoring 2x2+ 3x- 65?

factoring 2x^2+3x-65 would give me (2x+13)(x-5) so i'll cancel out and get (2x+13). So do I use direct substitution? f(5)= 2(5)+13 = 23?

Thanks for all the help . I appreciate it. I tried to make the colors different so it won't seem too confusing.

 

[HallsofIvy]

HallsofIvy, thanks for replying. I will try to post up what I did here.

23. What are the steps to finding

sqrt(3+6*x^2)/(2+5*x) as x goes to infinity?

i figured out the answer is sqrt(2) x sqrt (3) / 5 I don't get what I'm supposed to do to the numerator with the square roots.

What I did is turn that function disregarding the +2 and +3 so I got sqrt 6x^2 / 5x ... and I'm not suer what I do next...

Well, what is $\sqrt{6x^2}/5x$ for x positive?

To see that you can "disregard" the +2 and +3, divide both numerator and denominator by x. Since x is going to infinity it is not 0 so we certainly can do that and see that $\sqrt{2+ 6x^2}/(2+5x)= \sqrt{2/x^2+ 6}/(2/x+ 5)$. Now, as I said before, as x goes to infinity, 1/x goes to 0.

41. Find the Horizontal Asymptotes for

17x/(x^4+1)^1/4 . I'm pretty sure its 17/1, but for some reason, my teacher asks for two, and I thought rational functions only have one horizontal asymptote...

No, that is not necessarily so. "Horizontal asymptotes" occur is the graph approaches a horizontal line as x goes to infinity or negative infinity. What happens if x is a large negative number? For example, what is the value if x= -100000000?

What I did first is fix the denominator and make it become 17x / x+1 because I multiply the roots. I disregard the +1 and get 17x/x ... This has two horizontal asymptotes?
I thought because the ratio of the coefficient of 17x and x is 1 , n=m=1 therefore it is 17/1 ? please correct me if I'm wrong . if x is -10000 then it is neg infinity .
What happened to the fourth power and 1/4 power? (x⁴+ 1)^1/4 is NOT equal to x+1. In fact, even (x⁴)^1/4 is not equal to x- it is equal to |x|. Do you see why you get two different asymptotes?

And, why in the world would you think doing any calculation (where the denominator is not 0) would give "infinity"? If x= -10000, 17x/(x⁴+1)^1/4 is -170000/10000.00000000000025= -16.999999999999999575, not anywhere near "infinity"!

Is there any reason you keep writing "17/1" rather than "17"?

62. f(x)= 4x^3+13x^2+11x+24 / x+3 when x<-3
f(x)= 3x^2+3x+A when -3 less than or equal to x

What is A in order for it to be continuous at -3?

What is the definition of continuous? In particular, what are the limits as x goes to -3 from above and below?

continuous means there are no breaks , basically being able to draw it on a graph without lifting up pencil. also there is no discontinuity. as x goes to -3 when x<-3 it goes to 0?
I still don't understand what A has to be...

No, that is not the definition of continuity, it is a general property. The definition of "continuity of f(x) at x= a" requires three things: 1) f(a) exists; 2) $\lim_{x\rightarrow a} f(x)$ exists; 3) those are equal: $f(a)= \lim_{x\rightarrow a} f(x)$.

Of course, the limit exists if the two "one sided limits" exist and are the same. Often you can find the limit of f(x), as x goes to a, by finding f(a) because, often, the functions we work with are continous. If you try finding the limit of f(x) as x approaches -3 "from below" by evaluating (4x³+13x²+11x+24) /(x+3) by evaluating at x= -3, you get 0/0 so that doesn't work. But the fact that the numerator is 0 at x= -3 tells us that x-(-3)= x+ 3 is a factor. Find the other factor so you can cancel the "x+ 3" terms and find the limit.

The limit from above, $lim_{x\rightarrow -3^+} 3x^2+ 3x+ A$ is easy: it is a polynomial, so continuous so the limit is 3(-3)²+ 3(-3)+ A= 27- 9+ A= 18+ A.

In order that the given function be continuous at x= -3. Those two limits must be the same. Set them equal and solve for A.

79. f is continuous at (-inf, + inf)

f(y) = cy+3 range is (-inf,3)
f(y) = cy^2-3 range is (3,+inf)

what is C?

This makes no sense. First you must mean "on" (-inf, +inf), not "at". Second, do you mean "domain" rather than "range"? And, finally, the only way that would make sense is if the two domains are (-inf, 3] and [3,+inf). Assuming those are correct, what are the limits as x goes to 3 from above and below?

Sorry I did mean on, and yes it is the domain. the limits as x goes to 3 from above is -inf ?

Both "parts" of this function are polynomials and so the limits "from below" and "from above" can be found by evaluating them. What is c(3)+ 3 and c(3²)- 3? Set them equal and solve for c.

81. Let f(x) = {2x^2+3 x -65) / (x-5)

Show that f(x) has a removable discontinuity at x=5 and determine what value for f(5) would make f(x) continuous at x=5.
Must define f(5)=

You must define f(5) to be the limit as x approaches 5. f does not have a value at x= 5, given by that formula, because the denominator is 0 at x= 5. But in order to have a limit at x= 5, the numerator must also be ____. And what does that tell you about factoring 2x2+ 3x- 65?

factoring 2x^2+3x-65 would give me (2x+13)(x-5) so i'll cancel out and get (2x+13). So do I use direct substitution? f(5)= 2(5)+13 = 23?

Thanks for all the help . I appreciate it. I tried to make the colors different so it won't seem too confusing.

Yes, that's right. (2x+13)(x-5)/(x-5)= 2x+13 as long as x is not 5, and the limit as x approaches 5 depends only on what the value is for x close to 5, not equal to 5. Since (2x²+ 3x- 65)/(x-5)= 2x+ 13 for all x other than 5, they have the same limit at 5. and since 2x+ 13 is a polynomial, it is continuous and its limit is its value at x= 5.

 

[asdfsystema]

41. horizontal asymptotes are 17 and -17 then ?
62. i still don't really understand... what do you mean by x-(-3)= x+ 3 is a factor? i don't get the 4x^3+13x^2+11x+24/ x+3 when x<-3 part but I understand the other
81. the answer is 23 then?
79. the answer is 1 ?

another question is:
the vertical asymptote of -x^3/x-2 is x=2 right ?

sorry this is my first year learning calculus and I am not that familiar with the rules yet

 

[HallsofIvy]

41. horizontal asymptotes are 17 and -17 then ?

Yes, the horizotal asymptotes are 17 and -17.

62. i still don't really understand... what do you mean by x-(-3)= x+ 3 is a factor? i don't get the 4x^3+13x^2+11x+24/ x+3 when x<-3 part but I understand the other

If p(x) is a polynomial and p(a)= 0, then (x- a) is a factor of p(x): p(x)= (x-a)q(x) where q(x) is a polynomial of lower degree. In this case, since 4(-3)³+ 13(-3)²+ 11(-3)+ 23= 0, 4x³+ 13x²+ 11x+ 24= (x+3)q(x). You can find q(x) by dividing 4x³+ 13x²+ 11x+ 24 by x+3 (it was for precisely this purpose that "synthetic division" was developed). To put you out of your misery, 4x³+ 13x²+ 11x+ 24= (x+3)(4x²+ x+ 8). What is (4x³+ 13x²+ 11x+ 24)/(x+3)?

81. the answer is 23 then?

There are two parts to the question: the first is to show that there is a removeable discontinuity. You can do that by showing that the limit of the function, as x goes to 5, exists. Then the answer to the second part is that limit which is, yes, 23.

79. the answer is 1 ?

Yes, 9c- 3= 6c+ 3 gives c= 1.

another question is:
the vertical asymptote of -x^3/x-2 is x=2 right ?

Yes, that fraction is not defined at x= 2 so the graph cannot cross that vertical line. That is a vertical asymptote. (Notice that the graph cannot cross a vertical asymptote but can cross a horizontal asymptote. That is the because the definition of "function" does not allow two (x,y) points on the graph to have the same x value, but does allow two points with the same y value.)

sorry this is my first year learning calculus and I am not that familiar with the rules yet

 

[asdfsystema]

thanks i understand now . sorry for the trouble