Question

Please i need help in Pre cal

Answer 1

@MrNood

Answer 2

Do you know how to find the answer for A-E or do you need me to explain?

Answer 3

Start with question a. What is the domain of the function?

Answer 4

5x is the domain?

Answer 5

HI!!

Answer 6

denominator is \[x^2-5x+6=(x-2)(x-3)\] set that equal to zero, solve for \(x\)

Answer 7

you get \(x=2,x=3\) pretty much in your head, so the domain is all real numbers except those two

Answer 8

you could write that in a couple different ways
you could just say it
or you could say \[(-\infty,2)\cup (2,3)\cup (3,\infty)\]

Answer 9

the "points of discontinuity" are those two numbers, \(2\) and \(3\)

Answer 10

\[\frac{6-3x}{(x-2)(x-3)}=\frac{3(2-x)}{(x-2)(x-3)}\] and \[\frac{2-x}{x-2}=-1\] so the whole thing is just \[-\frac{1}{x-3}\]

Answer 11

that means \(2\) is a "removable discontinuity" because we just removed it

Answer 12

@Michele_Laino can you help continue please?

Answer 13

domain of a function is the set wherein the function does exist. So, using the results above, we can say that the function doesn't exist at points \(x=2\) and \(x=3\)

Answer 14

i got the domain but i dont know what the point of discontinuity is

Answer 15

so 2 and 3 are the domain

Answer 16

the points of discontinuity are the points at which the denominator becomes zero, namely \(x=2\) and \(x=3\).
So the domain of the function doesn't contain such points:

Answer 17

So zero is the domain

Answer 18

this is what it is?

Answer 19

domain is the rela line except the pints \(x=2\), \(x=3\), namely:
\[\huge domain = \left\{ {x \in \mathbb{R}|x \ne 2,\;x \ne 3} \right\}\]

Answer 20

that's right!

Answer 21

so my way isnt correct?

Answer 22

okay so now B

Answer 23

the points of discontinuity are the points \(x=2\) and \(x=3\)

Answer 24

okay now C

Answer 25

we can rewrite the function like below:
\[\Large \begin{gathered}
f\left( x \right) = \frac{{3\left( {2 - x} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)}} = \frac{{ - 3\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)}} = \hfill \\
\hfill \\
= \frac{{ - 3}}{{x - 3}} \hfill \\
\end{gathered} \]
as we can see the discontinuity at \(x=2\) disappears, so we say that \(x=2\) is a removable discontinuity

Answer 26

whereas \(x=3\) is still present, and we say that \(x=3\) is \(not\) a removable discontinuity

Answer 27

okay D!

Answer 28

x-intercepts are the values of \(x\) when \(f(x)=0\). So we have to solve this equation:
\[\Large f\left( x \right) = \frac{{3\left( {2 - x} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)}} = 0\]
which is equivalent to this one:

\[\Large \frac{{ - 3}}{{x - 3}} = 0\]

Answer 29

x=0 is the answer?

Answer 30

as we can see the solutions of equation:
\[\Large \frac{{ - 3}}{{x - 3}} = 0\]
are \(x= \pm \infty\), so we say that there are not x-intercepts

Answer 31

so that equation is the answer then

Answer 32

no, the answer is that there aren't x-intercepts

Answer 33

okay what about Y?

Answer 34

y intercepts are the value of y or f(x), when x=0, namely:
\[\Large f\left( 0 \right) = \frac{{3\left( {2 - 0} \right)}}{{\left( {0 - 2} \right)\left( {0 - 3} \right)}} = ...?\]

Answer 35

0

Answer 36

hint:
\[\Large f\left( 0 \right) = \frac{{3\left( {2 - 0} \right)}}{{\left( {0 - 2} \right)\left( {0 - 3} \right)}} = \frac{6}{{\left( { - 2} \right) \cdot \left( { - 3} \right)}} = ...?\]

Answer 37

9

Answer 38

it is y=1, since:
\[\Large f\left( 0 \right) = \frac{{3\left( {2 - 0} \right)}}{{\left( {0 - 2} \right)\left( {0 - 3} \right)}} = \frac{6}{{\left( { - 2} \right) \cdot \left( { - 3} \right)}} = \frac{6}{6} = 1\]
am I right?

Answer 39

so 1 is y

Answer 40

yes! we have one y-intercept only

Answer 41

okay thanks but do you know how to graph that last problem

Answer 42

do you know the theory of limits of functions, and the computation of first derivative of a function?

Answer 43

no

Answer 44

In that case I think that you can graph the function, by means of a graphyc calculator

Answer 45

for example try this calculator:
https://www.desmos.com/calculator

Answer 46

what do i put in?

Answer 47

https://www.desmos.com/calculator

Answer 48

please try to digit this information:
\[\Large y = \left( {x + 3} \right)/\left[ {\left( {x - 1} \right)\left( {x - 5} \right)} \right]\]
in the same way as I wrote

Answer 49

is this correct

Answer 50

that's the first function. Please you have to enter this:
\[\Large y = \left( {x + 3} \right)/\left[ {\left( {x - 1} \right)\left( {x - 5} \right)} \right]\]

Answer 51

ok! that's right!

Answer 52

thanks so much