ATTENTION DEAR MURICANS: This post contains some difficult math, please scroll to traps thread.
Is there a way (without calculus or graph) to get the range of this function?
The domain is easy, (-inf, -5]U[-3,inf) -[8]
but i can't figure out how the get the range without calc or graph methods.
Your domain is incorrect only 8 is out of the domain 3 and 5 result in a valid zero answer
Not your personal army/ tutor
Pay attention in lectur.
Ez pz factor that shit
(X-3)(x+5)
Yeah, that's why they are in a [ ] bracket
Found the amerifuck
Sorry, the correct domain is
(-inf, -5]U[3,inf) -[8]
-5 and 3 included, but not the numbers between
(-InfU8Uinf)
Same guy, to clarify, this factor is so that you know when the upper part in the sqrt is 0. If it changes from pos to neg, or vice versa, then it'll need to equal zero in between. Now we have the factored form, and we know that the part is 0 at -5 and 3. We can find out if it's pos or neg by putting in any given number between 3 and -5 (0). That gives us a neg number, so we know that the top DNE for (-5, 3).
Wrong, fgt. 0 returns Und
bump for self interest
THAT IS THE DOMAIN FUCKING AMERIFAG
OP WANTS THE FUCKING RANGE
THE RANGEEEEEEE!!!!
Dubs of truth
Haha, whoops, go fuck myself i guess.
Probably use 2nd derivative test, test for max and min values, then that's your range.
What class is this, for by the way?
It's still a solution even if it isn't on the real graph tho.
>The domain is easy, (-inf, -5]U[-3,inf) -[8]
your domain is wrong, R\{8} is the correct answer
also no you cannot know you have the range absolutely correct without calculus, but you can guess and check and put what looks like the right answer fairly easily but it won't be technically correct because you wouldn't have shown sound work
found the faggot euro who hasn't learned about i := sqrt(-1) yet
uh wtf the numbers in the middle most certainly are in the domain
for instance, f(-4):
sqrt((-4)^2 + 2(-4) - 15)/(-4-8)
sqrt(16-8-15)/(-12)
sqrt(-7)/(-12)
sqrt(7*-1)/(-12)
sqrt(7)i/(-12)
-sqrt(7)i/12
actually I mean C\{8} sorry
(i)^2 = -1 is correct, sqrt(-1) = i is a mistake so stop calling people faggot
the domain only includes REAL numbers
For large enough x, the top is the square root of x^2, so basically x, while the bottom is x, so as x->inf, f(x) -> 1.
Similarly, for x-> -inf, f(x)->-1.
The roots (-5,3) should just indicate where the graph crosses the x-axis. I believe there is a vertical assymptote at x=8, f(x)->-inf as x approaches 8 from the left and f(x) approaches +inf as x->8 from the right.
So your range is probably (-inf, -1)U(1, inf).
source: an american
Are you literally retarded?
btw that was just my first impression, i would actually check points around the interesting bits if i cared enough to get it right
>difficult math
I can't till if you're retarded and in your local community college's precalc or if you're b8.
...
>REAL
op didn't say that nigger
you're right, FAGGOT
acually he's right
assume i = sqrt(-1)
1 = sqrt(1) = sqrt(-1*-1) = sqrt(-1)sqrt(-1) = i*i = -1
clearly a contradiction, so the assumption is incorrect
Ahhhh amerifag detected!
Only real numbers you dumbfuck. It's so fucking obvious.
I agree, but the function can equal zero for real numbers. The range is more like (-inf,0]U(1,inf).
Winrar, but how do you prove it?
yeah missed that - basically the function gets chpoped off at (-5,0) and comes back at (3,0). After i checked points i also came up with (-inf, 0]U(1,inf).
Give me a minute.
>difficult math
Nigger please this is really low high school level.
being a calcfag, "proving it" to me means solving the limits as x->-inf, x->-5, x->3, x->8-, x->8+, and x->inf
but without doing that or graphing, not sure how you could "prove it"
check points, show values are growing etc?
insert infinity or zero into that scribberly shit. then solve... I can't even....
actually this is more complicated than most people will ever have to do in their lives
talk shit post solution faggot
lol math is dumb heh
If you are looking for a method to solve this without the graph, then you need to take derivatives and look at growth, decay, inflection, and extrema in the specific intervals. Which in the long run would just lead to you drawing a shitty graph. Some things in math do not have shortcuts. Especially calculus. All you are doing is solving trivial problems.
Here's how i came up with the basic behavior:
first, see for the reason that f(x) goes to -inf as x goes to - inf, and that f(x) goes to +inf as x-> +inf
then check some points, first on (-inf,-5]:
f(-100) = -.92
f(-10) = -.45
f(-6) = -.21
f(-5) = 0
seems like f(x) moves from an assymptote of -1 up to 0
check points on [3,8):
f(3) = 0
f(4) = -.75
f(5) = -1.49
f(6) = -2.87
f(7) = -6.93
seems like f(x) started at 0 and is getting more and more negative, going to -inf as x->8
now check (8, inf):
f(9) = 9.17
f(10) = 5.12
f(100) = 1.10
seems like f(x) goes to an assymptote of 1 as x->inf
but like said, checking all these points is basically graphing it, and the only other way is knowing where inflection points are which is calc