Question

functions

Answer 1

\(\large \color{black}{\begin{align} &\normalsize \text{Find the maximim value of the function}\hspace{.33em}\\~\\
& \dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

interval given ?

Answer 3

with out interval this maximum value diverges to infinity.

Answer 4

this question means that i have to minimize

\(\large \color{black}{\begin{align}
x^2-3x+2 \hspace{.33em}\\~\\
\end{align}}\)
right ?

Answer 5

should be, or els... i mean there is not maximum value here, if you are taking over \((-\infty,~+\infty)\)

Answer 6

wait

Answer 7

1/that ? or, just that ?

Answer 8

lol its


\(\large \color{black}{\begin{align}
\dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\
\end{align}}\)

Answer 9

\(\large\color{black}{ \displaystyle f(x)=\frac{1}{x^2-3x+2} }\)

and you want to find absolute minimum, right?

Answer 10

well, I can tell you that the limit as x approaches \(\pm\)infinity is going to be 0.

Answer 11

we can plot, I am guessing it is something like

Answer 12

i want to find the maximum value of this

\(\large \color{black}{\begin{align}
\dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\
\end{align}}\)

Answer 13

none

Answer 14

it is +infinity

Answer 15

but how ?

Answer 16

https://www.desmos.com/calculator

Answer 17

how i prove that with algebra

Answer 18

even the minimum doesn't exist.

Answer 19

with algebra, even with no calc ?

Answer 20

idk calculus

Answer 21

ic

Answer 22

lets c

Answer 23

\(\large\color{black}{ \displaystyle f(x)=\frac{1}{x^2-3x+2} }\)

it has vertical asymptotes, can you find them for me?

Answer 24

\(\large \color{black}{\begin{align}
x=2,\ 1 \hspace{.33em}\\~\\
\end{align}}\)

Answer 25

yes

Answer 26

and, the best thing I can see here is to use the idea of a limit to see where range tends to the close we get to x=1 and x=2.

I don't really know a purely algebraic proof for why this function has no max or min.

Answer 27

that has a max actually

Answer 28

yes?

Answer 29

https://www.desmos.com/calculator
keep scrollin up

Answer 30

on a certain interval, no doubt (unless it has a domain gap)

Answer 31

i think it attains one max
does not perhaps required any domain to get it or i think

Answer 32

your graph does not show up
see here
https://www.desmos.com/calculator

Answer 33

i don't think there is a max. that is due to an asymptote - two of them, there.

Answer 34

we can find the limit from the right and from the left as x approaches 1 and 2.

Answer 35

http://prntscr.com/7k4q1l

here see that bottom part of the graph

Answer 36

with calculus we can prove it has a max value
but @mathmath333 is looking for a different way

Answer 37

this graph goes further down and up if you scroll on desmos, but we can go ahead and take any limit that is closed from either side of any of the asymptotes you wish to choose.....

Answer 38

i mean here local max not abolute of course

Answer 39

how to find that local max

Answer 40

oh, but the question is asking for absolute doesn't it?

Answer 41

that is what i would think of course

Answer 42

first let's write that as 1/(x-1)(x-2)

Answer 43

i had thought of absolute too, just not sure about it!

Answer 44

yes the question is asking absolute, i just asked that of curiosity

Answer 45

this function is pretty weird. I wonder what would it model

Answer 46

i mean in real world