katherinesmith
  • katherinesmith
what is the approximate value of the function at x = -4? there is a graph.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
katherinesmith
  • katherinesmith
this is the graph
1 Attachment
zepdrix
  • zepdrix
So when we look over at x=-4, we try to figure out at what height the red line is located. Looks like it's between 1 and 2 yes? Maybe 1.75ish?
katherinesmith
  • katherinesmith
is it closer to 1.5 or 1.8? i'm assuming 1.8

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

zepdrix
  • zepdrix
Yah 1.8 sounds a little bit more accurate :)
zepdrix
  • zepdrix
It's probably 1.5 at around x=-6
katherinesmith
  • katherinesmith
i have another one so stay here!
katherinesmith
  • katherinesmith
what is the approximate value of the function at x = 2?
1 Attachment
zepdrix
  • zepdrix
Hmm, so our red line is below the x-axis at x=2, right? Take a guess! :)
katherinesmith
  • katherinesmith
either 0.25 or 0.50... positive or negative
zepdrix
  • zepdrix
Since it's below the x-axis, it will definitely be negative. It's between -1 and 0. Yah -0.25 seems like a good guess!
katherinesmith
  • katherinesmith
do you know how to do ones like : solve x/(x-2) < 2
zepdrix
  • zepdrix
\[\Large \frac{x}{x-2}\lt2\]Let's ummmm, multiply both sides by x-2,\[\Large x \lt 2(x-2)\]Understand that step? :o The (x-2)'s cancel on the left side.
katherinesmith
  • katherinesmith
yes.
katherinesmith
  • katherinesmith
so that leaves you with x < 2x - 4
katherinesmith
  • katherinesmith
then you put x on one side and get -x < - 4, then it becomes x > 4. correct answer?
zepdrix
  • zepdrix
Hmm one sec, yes that's part of our inequality. I think we should be getting another part also though. Lemme see what I messed up.
katherinesmith
  • katherinesmith
x > 4 could be a correct answer... i don't know why it wouldn't be. but let me know what you get.
zepdrix
  • zepdrix
Oh you have some multiple choice answers or no? :
katherinesmith
  • katherinesmith
no i don't.
zepdrix
  • zepdrix
oh :)
zepdrix
  • zepdrix
The problem is this... Remember what happens when we multiply/divide across an inequality? The inequality sign `flips` directions. Example:\[\Large -3x \lt 6\]Dividing both sides by -3 gives us,\[\Large x>-2\]See how the sign changed from `less than` to `greater than` ?
katherinesmith
  • katherinesmith
yes, which i did that.
zepdrix
  • zepdrix
So in our problem, I think it's a bad idea to multiply both sides by (x-2) as I did. We don't know whether or not that quantity is negative. So let's try something else :O
katherinesmith
  • katherinesmith
okay
zepdrix
  • zepdrix
\[\Large \frac{x}{x-2}\lt2\]Subtracting 2 from each side,\[\Large \frac{x}{x-2}-2<0\]Let's get a common denominator:\[\Large \frac{x}{x-2}-\frac{2(x-2)}{x-2}\lt0\]
zepdrix
  • zepdrix
So what does that simplify down to? :)\[\Large \frac{x-2(x-2)}{x-2}<0\]
zepdrix
  • zepdrix
Hmm I hope I'm doing this right, man I hate inequalities lol. Lemme check one of the smarty pants guys :P @SithsAndGiggles @satellite73
katherinesmith
  • katherinesmith
let me know. im still confused
anonymous
  • anonymous
I don't think putting the 2 on the other side is really necessary. I'd go with \[x<2(x-2)\] which gives \[x<2x-4\\ -x<-4\] Either way works, but I think this is simpler. ;)
katherinesmith
  • katherinesmith
so that leaves you with x > 4. because you divide by -x and flip the inequality.
zepdrix
  • zepdrix
Well if you wolfram this, it gives 2 intervals, x<2, x>4. So I was just getting a little confused :d
katherinesmith
  • katherinesmith
how did you get the two intervals? because thats' the answer i need
zepdrix
  • zepdrix
Because from this point, where I combined the fractions:\[\Large \frac{x-2(x-2)}{x-2}<0\]After we simplify, we get,\[\Large \frac{-x+4}{x-2}\lt0\]
zepdrix
  • zepdrix
If we look at the top and bottom separately (which I guess is what we're supposed to do),\[\Large -x+4\lt0 \qquad\qquad\qquad x-2\lt0\]
zepdrix
  • zepdrix
\[\Large -x+4\lt0 \qquad\to\qquad x>4\]\[\Large x-2\lt0 \quad\qquad\to\qquad x<2\]
zepdrix
  • zepdrix
That too confusing? :d Or was it back when we combined the fractions the part you got stuck on?
katherinesmith
  • katherinesmith
this all just hurts my head. lord jesus help me
zepdrix
  • zepdrix
XD
katherinesmith
  • katherinesmith
alright one more question
zepdrix
  • zepdrix
k c:
katherinesmith
  • katherinesmith
|dw:1377555967193:dw|
katherinesmith
  • katherinesmith
that's an = sign and then a - sign
zepdrix
  • zepdrix
Solve for x?
katherinesmith
  • katherinesmith
yes
zepdrix
  • zepdrix
Note:\[\Large x^2+x \qquad=\qquad x(x+1)\]
zepdrix
  • zepdrix
\[\Large \frac{x+5}{x(x+1)}=\frac{1}{x(x+1)}-\frac{x-6}{x+1}\]
katherinesmith
  • katherinesmith
then what the heck do you do
zepdrix
  • zepdrix
If we multiply both sides by (x+1), can you see what will happen? :o
katherinesmith
  • katherinesmith
please write it out for me
zepdrix
  • zepdrix
Lemme give you a simpler example really quick :) Maybe it will help it to click.\[\Large \frac{x}{3}=\frac{7}{3}\] If we multiply both sides by 3, what does it give us?
katherinesmith
  • katherinesmith
x = 7
zepdrix
  • zepdrix
good, all of the denominator 3's cancel with the 3's that we multiplied through by, right?
katherinesmith
  • katherinesmith
yes
katherinesmith
  • katherinesmith
\[\frac{ x + 5 }{ x } = \frac{ 1 }{ x } - x - 6\]
katherinesmith
  • katherinesmith
then what
zepdrix
  • zepdrix
One small thing to be careful of. We don't usually write it, but there is a pair of parenthesis implied when we have a negative in front of a fraction.\[\Large \frac{x+5}{x(x+1)}=\frac{1}{x(x+1)}-\frac{(x-6)}{x+1}\]
zepdrix
  • zepdrix
\[\Large \frac{ x + 5 }{ x } = \frac{ 1 }{ x } -(x - 6)\]
katherinesmith
  • katherinesmith
I got that. what next.
zepdrix
  • zepdrix
To finish getting rid of the fractions, let's multiply both sides by x.
zepdrix
  • zepdrix
Gives us something like this, yes? :)\[\Large (x+5)=1-x(x-6)\]
katherinesmith
  • katherinesmith
yes
zepdrix
  • zepdrix
From here, distribute your -x to each term in the brackets. And then try to get everything to one side, so the other side =0
zepdrix
  • zepdrix
crap i gotta go :U sorryyyy :C gotta go take care of something

Looking for something else?

Not the answer you are looking for? Search for more explanations.