Increasing/decreasing functions

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sigma

Junior Member
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Feb 19, 2006
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Find the interval(s) in which the function defined by \(\displaystyle \
\L\
f(x) = \frac{{4x + 4}}{{x^2 }}
\\) is increasing and in which interval(s) it is decreasing. Be sure to justify your answers. State the critical numbers and the coordinates of any relative maximum or minimum values that exists.

So if I find the derivitive of the function to be: \(\displaystyle \
\L\
f'(x) = \frac{{4(x^2 ) - (4x + 4)(2x)}}{{(x^2 )^2 }}
\\)

am I suppose to find the critical numbers after that? Can somebody help me out with that please? (if that is even what I'm suppose to do with this question here) I'm a little lost.
 
sigma said:
am I suppose to find the critical numbers after that?
Click to expand...
As per the usual procedure, yes, you need to find the critical numbers (the zeroes) of the first derivative: Set f' equal to zero, and solve.

Eliz.
 
If somebody could ever be so kind to verify this. I got 0 and 2 for critical numbers. The way I did that was this:
\(\displaystyle \
\L\
\begin{array}{l}
0 = \frac{{4(x^2 ) - (4x + 4)(2x)}}{{(x^2 )^2 }} \\
\to 0 = 4(x^2 ) - (4x + 4)(2x) \\
\to 0 = 4x^2 - 8x^2 + 8x \\
\to 0 = 4x(x - 2x + 2) \\
\to 0 = 4x(2 - x) \\
\to x = 0,2 \\
\end{array}
\\)
These are the critical numbers

Is that right?
 
sigma said:
If somebody could ever be so kind to verify this. I got 0 and 2 for critical numbers. The way I did that was this:
\(\displaystyle \
\L\
\begin{array}{l}
0 = \frac{{4(x^2 ) - (4x + 4)(2x)}}{{(x^2 )^2 }} \\
\to 0 = 4(x^2 ) - (4x + 4)(2x) \\
\to 0 = 4x^2 - 8x^2 + 8x \\
\to 0 = 4x(x - 2x + 2) \\
\to 0 = 4x(2 - x) \\
\to x = 0,2 \\
\end{array}
\\)
These are the critical numbers

Is that right?
Click to expand...

You made an algebraic mistake on the second "\(\displaystyle \to\)"
 
Almost, but you dropped a "minus" sign when you when from your third arrow line to your foruth arrow line.

Eliz.
 
Whoops. So the critical numbers would be 0 and -2 then?
 
I just want to see if I'm doing this right. Once I have the critical numbers, I make a sign diagram with the critical numbers plotted on to see where the intervals are increasing and decreasing? Here's my sign diagram.


Is this right? So it would be decreasing on \(\displaystyle \
\L\
\left( { - \infty , - 2} \right)
\\) and increasing on \(\displaystyle \
\L\
\left( { - 2,0} \right)
\\) and also increasing on \(\displaystyle \
\L\
\left( {0,\infty } \right)
\\)?

And the relative max is at \(\displaystyle \
\L\
\infty
\\) and min at -1? Now I'm confused. Whats the difference between absolute maximum and minimum and relative max and min? The way I was told to find these numbers is to plug the critical numbers back into the orginal function but that was always when they asked for the absolute max and min and they want relative in this case. What do I do? Please help!
 
Sigma, this is the derivative: \(\displaystyle \L
\frac{{ - 4\left( {x + 2} \right)}}{{x^3 }}\).

As you can plainly see, that expression as one zero: x=-2.
It is decreasing to the left of –2 and increasing on the right.
Therefore, x=-2 is a minimum!

What you wrote above cannot be. Infinity is not a number.
Thus it cannot be that either a maximum or a minimum can occur at infinity.
So x=-2 is the only critical point.
 
I don't understand how you got a derivative of \(\displaystyle \L
\frac{{ - 4\left( {x + 2} \right)}}{{x^3 }}\). My derivative was \(\displaystyle \
\L\
f'(x) = \frac{{4(x^2 ) - (4x + 4)(2x)}}{{(x^2 )^2 }}
\\)
and then did

\(\displaystyle \
\L\
\begin{array}{l}
0 = \frac{{4(x^2 ) - (4x + 4)(2x)}}{{(x^2 )^2 }} \\
\to 0 = 4(x^2 ) - (4x + 4)(2x) \\
\to 0 = 4x^2 - 8x^2 - 8x \\
\to 0 = 4x(x - 2x - 2) \\
\to 0 = 4x(-2 - x) \\
\to x = 0,-2 \\
\end{array}
\\)
and everybody so far was telling me that I was doing it right and had the correct critical numbers so now I'm confused. I also thought when doing the sign diagram that infinity can be a number because what happens if you have 0 as one of your critical numbers and you have a fractional function like this where you have to put it in the denominator if you have x's in the denominator and divide by zero? I look at other examples and they treat infinity as a number that if you say divide a negative number by zero, the number is negative infinity and vice versa if positive. This question is driving me nuts!
 
4x²-(4x+4)(2x) =
4x²-8x²-8x =
-4x(-x+2x+2) =
-4x(x+2)Divide by x^4 and you are there
 
SIGMA, there is really no need for you to be so confused by this problem.
It is a matter of algebra and definitions:
First the algebra: \(\displaystyle \L
\begin{array}{l}
f'(x) & = & \frac{{4x^2 - (4x + 4)(2x)}}{{\left( {x^2 } \right)^2 }} \\
& = & \frac{{4x^2 - 8x^2 - 8x}}{{\left( {x^4 } \right)}} \\
& = & \frac{{ - 4x^2 - 8x}}{{\left( {x^4 } \right)}} \\
& = & \frac{{ - 4x(x + 2)}}{{\left( {x^4 } \right)}} \\
& = & \frac{{ - 4(x + 2)}}{{\left( {x^3 } \right)}} \\
\end{array}\).
Note that dividing by x is legal because 0 is not in the domain of the original function.

That point about domain brings me to the second point: DEFINITIONS.
You need to look this up in your textbook.
However, the usual definition of critical point is:
A point c in the domain of f is a critical point of f if f’(c)=0 or f’ does not exist at x=c.
So the reason that x=0 is not a critical point in this question is that 0 in not in the domain of your function. (But that may not be true for the author(s) of your text, so check the textbook.)

As to the other question, infinity is never a number in any standard calculus text.
I think that you are misreading the textbook on this one.
 
Sorry. Just didn't see it before. So then I would assume the sign diagram looks more like this then?


and that its decreasing on\(\displaystyle \
\L\
( - \infty , - 2)
\\) and increasing on \(\displaystyle \
\L\
( - 2,\infty )
\\)? How do I find the relative max and min? Do I just take the critical number and put it into the original function? If thats the case, I only get -1 because its just one critical point but is that a max or min? I guess I can see -2 is obviously a min. So -1 would be a relative max then?
 
Now, think about this: f(-2) is the absolute minimum for the function.
The function decreases until x is –2 and then it increases to x=0.
Thus the lowest point is (-2,f(-2)), the absolute minimum.

Now note that if x>0 then f’(x)<0. Therefore it is decreasing for positive x.

Here what you have: f is decreasing on \(\displaystyle ( - \infty , - 2) \cup (0,\infty )\) and increasing on \(\displaystyle ( - 2,0)\).

But on \(\displaystyle (0, \infty)\) the function is positive and has no minimum
 
Ok I see now. How did you know though that if x>0, than f'(x)<0? Of course I would have known that to just use a value thats greater than zero and put it into f'(x) to see that but knowing to do that I wouldn't have known. A "0" would have to be on the sign diagram then I would have seen but because 0 can't be a critical number, I figured it doesn't go on the sign diagram. In that case, the sign diagram would look like this then.


and the relative min is at -2 and relative max is at 0 then? But I see now its the derivative function you use to find the negative and positive signs from the critical numbers.
 
There is NO MAX at 0!
There is no max for this function period!
It just does not happen.
Between -2 & 0 the function increases without bound, it does not reach any number.
The function increases without bound as x approaches 0 from the right.
BUT f(0) is not defined!
 
Actually, I just looked in my book and it says to find the relative max or min, take the critical numbers and put them into the orginal f(x) function. So for this, the relative min would be \(\displaystyle \
( - 2, - 1)
\\) then because if you put -2 into \(\displaystyle \
\frac{{4x + 4}}{{x^2 }}
\\) you get -1. So because there is no relative max because of what you said above, you just have a relative min at (-2,-1) then? Their looking for coordinates when it comes to relative max or min so that's why their listed in a pair like that. If so, then that answers my question period. I just always thought you always had to have both a relative max and a relative min. These types of questions are very new to me (I just learned them 3 days ago.)
 
Yes, that seems right? I do get it now but I just want to make sure.
 
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