Finding Critical Points & Extrema (Local & Absolute), & Trends on Open Intervals

**SwatchAndGo** · November 2nd 2012, 04:37 PM

Hi.

First post on the forum, but regardless of which, need some assistance.

I have the following equation: $f(x)=\frac{x}{x+3}$

My purpose is, to quote the text:

Find the critical numbers of f (if any). Find the open intervals on which the function is increasing or decreasing and locate all relative extrema. Use a graphing utility to confirm your results.

I know I must take the derivative of the equation to find the cp's, however, I don't know exactly how to do it "the shortcut way" (where the fraction is made into a negative exponential power).

When I use the quotient rule, I get the following: $f'(x)=\frac{3}{(x+3)^2}$

After that, I'm not sure how to factor the equation to get the cp's. I know that's done by setting it equal to zero; factoring is where I'm

.

**Prove It** · November 2nd 2012, 04:39 PM

Originally Posted by SwatchAndGo

Hi.

First post on the forum, but regardless of which, need some assistance.

I have the following equation: $f(x)=\frac{x}{x+3}$

My purpose is, to quote the text:

I know I must take the derivative of the equation to find the cp's, however, I don't know exactly how to do it "the shortcut way" (where the fraction is made into a negative exponential power).

When I use the quotient rule, I get the following: $f'(x)=\frac{3}{(x+3)^2}$

After that, I'm not sure how to factor the equation to get the cp's. I know that's done by setting it equal to zero; factoring is where I'm

.

The equation $\displaystyle \begin{align*} \frac{3}{(x + 3)^2} = 0 \end{align*}$ does not have any solutions, so there are not any critical points.

**SwatchAndGo** · November 2nd 2012, 05:51 PM

Alright. How did you get to that conclusion? Or should that be just assumed at this point?

Also, I'm working on the next problem on the set that was assigned to me:

$f(x)=\frac{x+4}{x^2}$ => Quotient Rule => $f'(x)=\frac{(x^2)(1)-(x+4)(2x)}{(x^2)^2}$ => Simplify => $f'(x)=\frac{x^2-2x^2-8x}{x^4}$

So far, I came up with this as its derivative (using the quotient rule):

$f'(x)\frac{-x(x+8)}{x^4}$

Not exactly sure how to solve for zero (if there are any solutions)

**MarkFL** · November 2nd 2012, 06:15 PM

In your first problem, you correctly differentiated to obtain:

$f'(x)=\frac{3}{(x+3)^2}$

Now, the only way for this to equal zero is when the numerator is zero, but it is a non-zero constant, so this can never happen for any $x$ .

You do have a critical value from the denominator of $x=-3$ , but since this is a root of even multiplicity the sign of the derivative will not change across this point. So we know the function is increasing on:

$(-\infty,-3)\,\cup\,(-3,\infty)$

We know the original function has a vertical asymptote at $x=-3$ and a horizontal asymptote at $y=1$ .

For your second problem, you may simplify your derivative by dividing out the common factor of $x$ , i.e.:

$f'(x)=-\frac{x+8}{x^3}$

Now, equate the numerator to zero to find stationary point(s) and equate the denominator to zero to find point(s) where the sign of the derivative may change. Since the denominator has a root of odd multiplicity we can expect the derivative will change sign across this point. Once you have identified your critical numbers, divide the number line at these points and test the resulting intervals.

edit: Can you identify the asymptotes of the original function?

**SwatchAndGo** · November 2nd 2012, 06:41 PM

Ok, so even though $f(x)=\frac{x}{x+3}$ has a vertical asymptote at $-3$ , the function's slope is still positive across the asymptote.

V.A. @ x = 0, H.A. @ y = 4? (For the original equation)

I didn't understand where you got $-\frac{x+8}{x^3}$ from until I understood you canceled out the x: $\frac{-x(x+8)}{x^4}$ . But don't the top terms $x$ and $8$ tie to each other through addition and therefore you can't factor out x from one without the other?

I'd get something along these lines: $-\frac{1}{x^2}+\frac{8}{x^3}$ , which doesn't exactly make it obvious where to go from there.

What do stationary points refer to?

There's basically only one cp that can be identified, which is at 0, a V.A.

**MarkFL** · November 2nd 2012, 06:56 PM

No, the slope isn't positive across $x=-3$ , but it is positive on either side. That's why I gave two intervals, to exclude $x=-3$ .

For the second problem, the vertical asymptote is $x=0$ , but the horizontal asymptote is not $y=4$ . The degree in the denominator is greater than the degree in the numerator, so the horizontal asymptote is $y=0$ . You may confirm this by considering:

$\lim_{x\to\pm\infty}\frac{x+4}{x^2}=\lim_{x\to\pm \infty}\left(\frac{1}{x}+\frac{4}{x^2} \right)=0$

In the second problem, when you go to simplify the derivative, since $x$ is a factor in both the numerator and denominator, we can divide it out.

Stationary points refer to where the derivative is equal to zero. If the derivative changes sign across this point, then it is a relative extremum.

There is another critical value, which you can find by equating the numerator to zero:

$x+8=0$

$x=-8$

So test the sign of the derivative on the intervals $(-\infty,-8),\,(-8,0),\,(0,\infty)$ to find where the original function is increasing/decreasing.

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