Absolute maximum and Absolute minimum

[ZeroXz]

Hi,

Please I need some help.

I don't understand how come the answer for this question is no absolute max, no absolute min.

f(X) = 3x^2 / (x - 3) on [0, 4] \ {3}

Can someone please help me out?

I set f'(x) = 0 and got x = 0
Then, f(0) = 0

For end points,
f(0) = 0
f(4) = 48

I got absolute max = 48, absolute min = 0 but it's wrong and I don't understand.

Please help, thanks a lot!!

 

[tkhunny]

Around the vertical asymptote at x = 3, the function is unbounded in both the positive and negative directions.

 

[soroban]

Hello, ZeroXz!

I don't understand how come the answer for this question is: no abs.max, no abs.min.

. . \(\displaystyle f(x) \:=\: \frac{3x^2}{x - 3}\:\text{ on }[0, 4]\)


\(\displaystyle \text{I set }f'(x) = 0\text{ and got }x = 0\)
\(\displaystyle \text{Then: } f(0) = 0\)

\(\displaystyle \text{For end points, }\:f(0) = 0,\;f(4) = 48\)

\(\displaystyle \text{I got absolute max = 48, absolute min = 0, but it's wrong and I don't understand.}\)

You overlooked the vertical asymptote at \(\displaystyle x = 3\)

The graph looks like this:

            |       :
            |       :@
            |       :
            |       : @              *
            |       :  @           *
            |       :    @      *
            |       :       *
            |       :
  ----------@-------:----+---------------
        *   |  @    :3   4
     *      |    @  :
   *        |     @ :
            |       :
            |      @:
            |       :

You found a local maximum at (0, 0) and a local minimum at (4, 48)

But there is no absolute maximum or absolute minimum.

 

[ZeroXz]

Hi,

But I thought x = 3 is excluded from the domain [0, 4] \ {3}

How come I must consider x = 3 ??

 

[tkhunny]

Read VERY CAREFULLY what I wrote.

I did NOT say, "The function is unbounded AT x = 3." You are correct. x = 3 is not in the Domain.