Question

f(x)=(6x−2)/(x+3) find the min and max

Answer 1

Find f'(x) (with the product or quotient rule) and let it equal zero, solve for x.

You might want to check the sign of f''(x) again too, to check it's positive (minimum) or negative (maximum) or zero (saddle point, neither max or min).

Answer 2

The graph is increasing everywhere and has an infinite discontinuity at x = -3. So it has no min/max over it's Domain.

Answer 3

i just need someone to help me see if im doing ok for my derivative i got 20/(x+3)^2

Answer 4

Correct!

And you can see, as @genius12 said, the f'(x) will always be positive.

Answer 5

oh i had a question how would i know when its decreasing then

Answer 6

It'll be decreasing if f'(x) is ever negative. Can it be, based on what you found for f'?

Answer 7

If f'(x) > 0, function is increasing.
If f'(x) < 0, function is decreasing.

Answer 8

to write my intervals then how would i say its decreasing from -inf,inf

Answer 9

You could say it like this:
\[\bf f'(x) < 0 \ for \ x \in (-\infty,\infty)\]

Answer 10

But in this case, our f'(x) > 0 for all x's.

Answer 11

i got confuse so for this function then its not decreasing

Answer 12

It never decreases, since this can never be negative
\[\Large f'(x) = \frac{ 20 }{ )x+3)^2}\]

Can you plug in any x to make it negative? The squared part in the denominator is always positive.

Answer 13

i believe what you guys are saying but my professor apparently sees that its decreasing somewhere and wants me to write the interval

Answer 14

He's either misprinted the question then, or has just made a mistake.

You go to UCR right? One of the kids I tutor, his dad is a professor at UCR, but for physics, so it's probably not him...

Answer 15

For this function, like agent stated, the derivative is the following:\[f'(x)=\frac{ 20 }{ (x+3)^2 }\]This derivative is always positive no matter what value of x you plug in. This means that f'(x) > 0 for all x's in the domain, which means that the function is always increasing.

Answer 16

yeah probably well thanks anyway ill just leave it blank

Answer 17

im not sure if you guys can still see this but i had a question is my critical point 20

Answer 18

In this case, there aren't any critical points, in terms of the slope of the line, since f'(x) is never zero.

There's a vertical asymptote at x=-3, since if you put x=-3 into the original equation \[\Large f(x)=\frac{6x−2 }{x+3}\]
the denominator will be zero.

Answer 19

hmm so then how am i suppose to find my max and min i was trying to do the first derivative test thats why

Answer 20

In this case you can't find a max/min turning point.

The max of the graph is just positive infinity, the minimum is negative infinity - you can see this if you zoom out on the graph:
https://www.google.com/search?q=(6x-2)%2F(x%2B3)&aq=f&oq=(6x-2)%2F(x%2B3)&aqs=chrome.0.57j0l3j62l2.1809j0&sourceid=chrome&ie=UTF-8

also if you plug x=-3 into the derivative \[\large f'(x)=\frac{ 20 }{ (x+3)^2 }\] the slope is undefined (ie it's a vertical line for the slope)

Answer 21

Are you sure the function is\[\Large f(x)=(6x−2)/(x+3)\] ?

Answer 22

yeah i guess ill just leave them blank but there is concavity right

Answer 23

Yes, the concavity you can find by finding f''(x). Then you just have to see where it's positive and negative, and f'' should be positive for x<-3, and negative for x>-3

Answer 24

okay thanks again :)