asdfsystema said:
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? (x
4+ 1)
1/4 is NOT equal to x+1. In fact, even (x
4)
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/(x
4+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 (4x
3+13x
2+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(3
2)- 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 (2x
2+ 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.
