Find the domain of a function

[desert gazelle]

stuck here at picture 4

[vapid locust]

In picture 4 you just have to consider more cases

[desert gazelle]

my friends are now telling me to draw a chart like this:

are they correct?

btw, that's not a solution yet

[vapid locust]

The first time i have seen something like this, I don't know if this is valid but my way is just considering more cases

[desert gazelle]

okay, then I have those 4 cases from the pic above the last one

I have a question

[vapid locust]

What

[desert gazelle]

Can I leave it in the form of quadratic equation?

This is the original assignment

and I have to find the domain

[vapid locust]

You can but it doesn't help because if you want to split the log you are going to have to consider when all of them are positive

[desert gazelle]

can we start from this image:

I found what conditions I need

one is that x+1 mustn't equal 0

the other that the whole fraction must be higher than 0

let's start from there

[vapid locust]

True

[desert gazelle]

okay, so my solution mustn't include -1

now moving onto the fraction

how do we solve it?

we need to find all x values that will result in a positive fraction?

[vapid locust]

There's probably easier ways, but I am going to give what I can think of now

[vapid locust]

Yes

[desert gazelle]

alright

[vapid locust]

By the way, this is if you want to split the log you can find the limitations then see what limitations you get and use the x value that include all positive

[desert gazelle]

how to split it?

[vapid locust]

Like for example log(20-x)+log(x+10) then you say x+10>0 and 20-x>0

[vapid locust]

Using log properties log(xy)=logx+logy and log(x/y)=logx-logy

[desert gazelle]

Let's do that

[vapid locust]

I think that would be easier than the first way

[desert gazelle]

log(x/y)=logx-logy

[vapid locust]

Yes

[desert gazelle]

How do you solve this? We are not allowed calculators on exam

[vapid locust]

Wait maybe this way doesn't work because it doesn't account for when both are negative and they cancel

[desert gazelle]

they can be both negative or both positive

we must get a positive number

right?

[vapid locust]

Sorry I think we should go back to the old way

[vapid locust]

Yes

[desert gazelle]

ok

[vapid locust]

Ok so let's consider the inside of the log bigger than zero

[desert gazelle]

ok

can we try putting in various numbers

instead of x

1 won't do

-1 won't do

[vapid locust]

I don't think that works going back to the fourth picture we can consider when like (x-1)(x-2)>0 then say it's either x-1>0 or x-1<0

[desert gazelle]

but why transfer it from quadratic to two binomails?

[vapid locust]

So we can find the solution for the inequity

[desert gazelle]

you can leave it quadratic and say xsq -3x +2 >0

[vapid locust]

You can but simplification after that is not easy because it's an inequality

[desert gazelle]

nominator and denominator must both have same presign

so we make into 2 cases like this

then we solve

find the interval

[vapid locust]

Yes also you have to consider two cases in every case

[desert gazelle]

maybe

2 and above works I think in first case

sorry, above 2

so that's one

next we go below -1

everything lower than -1 works too

[vapid locust]

We say x-1>0 when x-2>0 or x-1<0 when x-2<0

[desert gazelle]

you confused me

this is good I think

now second case

[vapid locust]

I don't know how to say it, x-2 could be negative when x-1 is positive so you have to consider both cases when x-1 and x-2 are positive or negative

[desert gazelle]

yes I know

we are not finished

we must find intersection

[vapid locust]

Oh so you considering when both are positive in first case?

[vapid locust]

I don't how that works but if it does use it I don't know

[desert gazelle]

we combine the two intervals

Photomath solve it

I must go to dinner

[vapid locust]

Ok use it

[desert gazelle]

I come later

see you

thanks

[vapid locust]

You can check after solving on wolfram alpha or look in a graphing calculator

[delicate blaze]

I would approach this by solving the inequality:

$\frac{x^2 - 3x + 2}{x+1} > 0$

It is tempting to multiply by $(x+1)$ but, even assuming $x\neq -1$,

you don't know what $x$ is and so you don't know if $(x+1)$ is

positive or negative, and so you can't know whether to retain or

flip the direction of the inequality sign.

This problem is avoided by multiplying by $(x+1)^2$, which can't be negative:

$\frac{(x-1)(x-2)}{x+1} \times (x+1)^2 > 0 \times (x+1)^2$ provided $x \neq -1$

$(x-1)(x-2)(x+1) > 0$

Sketching this positive cubic functionwith single roots at $x= -1, 1, 2$,

it is clear that it is above the $x$-axis for $x \in (-1,,1)\cup(2,,\infty)$.