Question

How do I put this in my graphing calculator and get the answer: (-3.449, -11.899)
f(x)= (x^2 - 5x)/(x+1), x<-1

Answer 1

\[f(x)=\frac{ x^2-5x }{ x+1 }, x<-1\]

Answer 2

More than one point is the solution to that, so what is the original question associated with that function?

Otherwise, no one will have a clue how to get that answer. Help us help you

Answer 3

Find the maximum of this function

Answer 4

To find the max value of the function

1. Find f'(x).
2. Set f'(x) = 0.
3. Solve for x using your calc.
4. Plug x back into the expression for f(x)
5. Evaluate the expression using your calc.

Answer 5

for the derivative i got \[\frac{ x^2 +2x - 5 }{ (x+1)^2 } = 0\]

Answer 6

so in order to find the zeros i would have to get the quadratic formula for this equation?

Answer 7

Make sure to multiply both sides of the equation by (x+1)^2 first

x^2 + 2x - 5 = 0

Answer 8

so when i do that i'll just be left off with x^2 +2x - 5, then i use the quadratic formula to find the zeros?

Answer 9

@Hero

Answer 10

how did u get the x^2 +2x +4 part?

Answer 11

Actually, hang on. I made a mistake

Answer 12

"Find the maximum to these functions" the answer my teacher gave me was (-3.449, -11.899)

Answer 13

he used the graphing calculator to get the answer but when I used the graphing calc. I got a way different answer than he did x-x

Answer 14

Bro, I just told you I made a mistake. Will you let me correct it? PLEASE

Answer 15

It was not necessary to repeat THE SAME THING TWICE

Answer 16

sorry

Answer 17

x^2 + 2x = 5
x^2 + 2x + 1 = 5 + 1
(x + 1)^2 = 6
x + 1 = ± √(6)
x = ± √(6) - 1
x = √(6) - 1
x = -√(6) - 1

Answer 18

There. That should work

Obviously you will only be choosing the positive point

Answer 19

how did uget the x^2 +2x +1 part

Answer 20

You add (b/2)^2 to both sides. That's what is called completing the square.

Answer 21

so since b = 2,

(2/2)^2 = 1

Answer 22

so you add 1 to both sides

Answer 23

Sorry, you choose the negative x value, not the positive one

Answer 24

so i dint have to use the quadratic formula for this?

Answer 25

There's more than one way to solve quadratic equations. I suppose your teacher never told you that.

Answer 26

well when i did the quadratic formula for this I got \[\frac{ -2 \pm \sqrt{24}}{ 2}\] is that correct?

Answer 27

Yes, that is correct because if you keep reducing you end up with the same thing:

\[\frac{-2 \pm \sqrt{(4)(6)}}{2}\]
\[\frac{-2 \pm \sqrt{4}\sqrt{6}}{2}\]
\[\frac{-2 \pm 2\sqrt{6}}{2}\]
\[2\left(\frac{-1 \pm \sqrt{6}}{2}\right)\]
\[-1 \pm \sqrt{6}\]
\[\pm \sqrt{6} -1\]

Answer 28

so i would always choose the answer that is negative for these kinds of problems? "Finding the maximum to these functions"

Answer 29

Not always

Answer 30

For this particular problem, it was stated that x < -1

Answer 31

oh okay ty :DDD

Answer 32

but my teacher used the graphing calculator for this, can u show me how he did b/c when I try to plug in the equation on my calc. i get a weird graph and can't find a maximum on there

Answer 33

No, I cannot show you how he did it. I'm not your teacher and I don't have his graphing calc. All I can do is tell you how I would do it. Besides I use TI-Nspire Graphing Calc. Your teacher probably uses TI-89 or something older.

Answer 34

oh. ok ty ^^

Answer 35

can u hep me with another problem

Answer 36

All you have to do is just post your question. Someone will help. But close this question first. Then post your question as a new one.