Help calculus question

[hamza4best]

So this is the function f(x)=(x+4)/(x^2+5x+4)

Can anyone confirm that I got the right answers

1. Domain: All real nos except -4 and -1

2. Range: (-oo,+oo)

3. For critical points I get x=-4 and x=-1 but since the function is not defined at these points so no critical points right?

4. The increasing/decreasing interval for the function will be decreasing entirely

5. At x=-4 the function will be continuous or not?

6. The inflection point I got was x=-1 but since again the function is also undefined at this point so there will be no inflection point

7. For concavity the function will have concave down decreasing from (-oo,-1) and concave up decreasing from (-1,+oo)

 

[pappus]

So this is the function f(x)=(x+4)/(x^2+5x+4)

You certainely noticed that \(\displaystyle \displaystyle{f(x)=\frac{x+4}{x^2+5x+4} = \frac1{x+1}~,~x\ne-4}\)

Can anyone confirm that I got the right answers

1. Domain: All real nos except -4 and -1 OK

2. Range: (-oo,+oo) OK

3. For critical points I get x=-4 and x=-1 but since the function is not defined at these points so no critical points right?

If you use the simplified term of the function those considerations are not necessary

4. The increasing/decreasing interval for the function will be decreasing entirely

5. At x=-4 the function will be continuous or not?

Use the simplified term and some limits to answer this question.

6. The inflection point I got was x=-1 but since again the function is also undefined at this point so there will be no inflection point

7. For concavity the function will have concave down decreasing from (-oo,-1) and concave up decreasing from (-1,+oo)

 

[soroban]

Hello, hamza4best!

Given: .\(\displaystyle f(x)\:=\:\dfrac{x+4}{x^2+5x+4}\)

Can anyone confirm that I got the right answers?

1. Domain: All real nos except -4 and -1 . Right!


2. Range: (-oo,+oo) . Not quite . . .

We have: .\(\displaystyle f(x) \:=\:\dfrac{x+4}{(x+1)(x+4)} \:=\:\dfrac{1}{x+1}\;\text{ for }x \ne \text{-}4\)

The graph looks like this:

                :    |
                :*   |
                :    |
                : *  |
                :  * |
                :    *
                :    |  *
                :    |      *
  - - - - - - - : - -+- - - - - - -
    *          -1    |
        o       :    |
           *    :    |
             *  :    |
              * :    |
                :    |
               *:    |
                :    |

It is the graph of \(\displaystyle f(x) = \dfrac{1}{x+1}\) .with a "hole" at \(\displaystyle (\text{-}4,\,\text{-}\frac{1}{3})\)

Therefore, the range is: .\(\displaystyle (\text{-}\infty,\,\text{-}\tfrac{1}{3})\,\cup\,(\text{-}\tfrac{1}{3},\,\infty) \)



3. For critical points I get x=-4 and x=-1 but since the function
is not defined at these points so no critical points, right?

You are right ... for the wrong reason.

Critical values occur when \(\displaystyle f'(x) = 0.\)

Since \(\displaystyle f'(x) \,=\,\text{-}\dfrac{1}{(x+1)^2}\), the derivative can never equal zero.
And that is why there are no critical values.



4. The increasing/decreasing interval for the function will be decreasing entirely

Correct . . . the derivative \(\displaystyle \text{-}\frac{1}{(x+1)^2}\) is always negative.
Hence, the function is always decreasing.



5. At x=-4 the function will be continuous or not?

At \(\displaystyle x = \text{-}4\), the graph does not exist.
The function is not continuous there.



6. The inflection point I got was x=-1.

Inflection points occur where \(\displaystyle f''(x) = 0\)

Since \(\displaystyle f''(x) = \dfrac{2}{(x+1)^3}\), the second derivative can never equal zero.
There are no inflection points.



7. For concavity the function will be concave down decreasing on (-oo,-1)
and concave up decreasing on (-1,+oo)

Correct!

 

[hamza4best]

Hello, hamza4best!1. Domain: All real nos except -4 and -1 . Right!2. Range: (-oo,+oo) . Not quite . . .We have: .\(\displaystyle f(x) \:=\:\dfrac{x+4}{(x+1)(x+4)} \:=\:\dfrac{1}{x+1}\;\text{ for }x \ne \text{-}4\)The graph looks like this:

                :    |                :*   |                :    |                : *  |                :  * |                :    *                :    |  *                :    |      *  - - - - - - - : - -+- - - - - - -    *          -1    |        o       :    |           *    :    |             *  :    |              * :    |                :    |               *:    |                :    |

It is the graph of \(\displaystyle f(x) = \dfrac{1}{x+1}\) .with a "hole" at \(\displaystyle (\text{-}4,\,\text{-}\frac{1}{3})\)Therefore, the range is: .\(\displaystyle (\text{-}\infty,\,\text{-}\tfrac{1}{3})\,\cup\,(\text{-}\tfrac{1}{3},\,\infty) \)3. For critical points I get x=-4 and x=-1 but since the function is not defined at these points so no critical points, right?You are right ... for the wrong reason.Critical values occur when \(\displaystyle f'(x) = 0.\)Since \(\displaystyle f'(x) \,=\,\text{-}\dfrac{1}{(x+1)^2}\), the derivative can never equal zero.And that is why there are no critical values.4. The increasing/decreasing interval for the function will be decreasing entirelyCorrect . . . the derivative \(\displaystyle \text{-}\frac{1}{(x+1)^2}\) is always negative.Hence, the function is always decreasing.5. At x=-4 the function will be continuous or not?At \(\displaystyle x = \text{-}4\), the graph does not exist.The function is not continuous there.6. The inflection point I got was x=-1.Inflection points occur where \(\displaystyle f''(x) = 0\)Since \(\displaystyle f''(x) = \dfrac{2}{(x+1)^3}\), the second derivative can never equal zero.There are no inflection points.7. For concavity the function will be concave down decreasing on (-oo,-1)and concave up decreasing on (-1,+oo)Correct!

Few things which I'm not getting1. If we simplify the function to 1/(x+1)then at x=-4 we get -0.333 but if we put it without the simplification we get undefined why?2. For continuity we have to put the values which are close to -4 i.e x approaches -4so for left hand limit if we put -4.00001 and right hand limit we put .3.99999 we get the answer somewhere close to -0.3333 so shouldn't it be continuous then at x=-4?

 

[srmichael]

It is true that there is a hole at (-4, -1/3), but also you must realize that the limit of this function as x goes to positive or negative infinity is 0 so the range is actually:

(-∞, -1/3) U (-1/3, 0) U (0, ∞)

 

[JeffM]

Few things which I'm not getting1. If we simplify the function to 1/(x+1)then at x=-4 we get -0.333 but if we put it without the simplification we get undefined why?2. For continuity we have to put the values which are close to -4 i.e x approaches -4so for left hand limit if we put -4.00001 and right hand limit we put .3.99999 we get the answer somewhere close to -0.3333 so shouldn't it be continuous then at x=-4?

These are good questions.

\(\displaystyle f(x) = \dfrac{x + 4}{x^2 + 5x + 4} \implies f(x)\ does\ not\ exist\ if\ x = - 1\ or\ x = - 4.\)

Therefore you cannot say what f(x) equals at x = - 1 or x = - 4. It does not equal anything; it is not there. The fact that f(x) = g(x) at an infinite number of x values does not overcome the fact that f(x), as it has been defined, does not equal g(- 4) because there is no such thing as f(-4) as it has been defined. We are being super careful about definitions. If we want f(x) to exist at x = - 4, we must define f(x) differently.

\(\displaystyle f(x) = \dfrac{x + 4}{x^2 + 5x + 4}\ if\ x \ne - 4;\ f(-4) = - \dfrac{1}{3}\) works.

What also works is \(\displaystyle f(x) = \dfrac{1}{x + 1}.\)

But we did not define f(x) in those ways. For some reason, we wanted to exclude x = - 4 from consideration. So we defined f(x) in such a way to accomplish that task.

Does this help with your first question?

As for your second question, it too is a matter of being very careful about definitions.

For f(x) is defined to be continuous at x = a if and only if,

\(\displaystyle f(a)\ exists,\ \displaystyle \lim_{x \rightarrow a}f(x)\ exists,\ and\ f(a) = \lim_{x \rightarrow a}f(x).\)

You are correct that the limit exists, but the other two conditions are not met because f(-4) does not exist and so cannot equal the limit.

Clear now?

 

[hamza4best]

These are good questions.

\(\displaystyle f(x) = \dfrac{x + 4}{x^2 + 5x + 4} \implies f(x)\ does\ not\ exist\ if\ x = - 1\ or\ x = - 4.\)

Therefore you cannot say what f(x) equals at x = - 1 or x = - 4. It does not equal anything; it is not there. The fact that f(x) = g(x) at an infinite number of x values does not overcome the fact that f(x), as it has been defined, does not equal g(- 4) because there is no such thing as f(-4) as it has been defined. We are being super careful about definitions. If we want f(x) to exist at x = - 4, we must define f(x) differently.

\(\displaystyle f(x) = \dfrac{x + 4}{x^2 + 5x + 4}\ if\ x \ne - 4;\ f(-4) = - \dfrac{1}{3}\) works.

What also works is \(\displaystyle f(x) = \dfrac{1}{x + 1}.\)

But we did not define f(x) in those ways. For some reason, we wanted to exclude x = - 4 from consideration. So we defined f(x) in such a way to accomplish that task.

Does this help with your first question?

As for your second question, it too is a matter of being very careful about definitions.

For f(x) is defined to be continuous at x = a if and only if,

\(\displaystyle f(a)\ exists,\ \displaystyle \lim_{x \rightarrow a}f(x)\ exists,\ and\ f(a) = \lim_{x \rightarrow a}f(x).\)

You are correct that the limit exists, but the other two conditions are not met because f(-4) does not exist and so cannot equal the limit.

Clear now?

I think I got it
What you're saying is that If we simplify the f(x) then it becomes another function g(x) so the behaviour of g(x) will be different from the original function in the sense that the values which we can't input in f(x) can be entered into g(x)?
As for my second question
How do I write the discontinuity in mathematical form
f(-4)=undefined
limit(x approaches -4)=undefined
so f(-4) not equal to limit(x approaches -4)
will that be correct?

Lastly for the range(we see the output) so in that case as we can't input -4 and -1 so we wont be able to include the output of these two values?

 

[JeffM]

What you're saying is that If we simplify the f(x) then it becomes another function g(x) so the behaviour of g(x) will be different from the original function in the sense that the values which we can't input in f(x) can be entered into g(x)?

Yes. If f(x) and g(x) have different domains, then they are completely different functions. If - 4 is not in the domain of f(x), the fact that - 4 is in the domain of g(x) does not magically change the fact that - 4 is not in the domain of f(x).

As for my second question
How do I write the discontinuity in mathematical form
f(-4)=undefined

One way is: \(\displaystyle f(- 4)\ \nexists.\)

A better way is to use the the interval notation shown in Sir Michael's post above:

\(\displaystyle Domain\ of\ f(x)\ is\ (- \infty,\ - 4)\ \bigcup\ (- 4,\ -1)\ \bigcup\ (-1,\ \infty).\)

Clarifying edit: The left and right round brackets around the intervals indicate that NONE of the delimiting numbers are included in the intervals. So this notation tells you what is included in the domain, namely every real number except - 1 and - 4, and what is excluded, namely - 1 and - 4 and every number that is not real. Much more informative than just saying - 4 is not in the domain.

I think I got it limit(x approaches -4)=undefined
so f(-4) not equal to limit(x approaches -4)
will that be correct?

Not quite. As you yourself showed, the limit of f(x) as approaches - 4 DOES exist even though the function itself does not exist at x = - 4. Consequently, they cannot be equal: something is not equal to nothing.

Lastly for the range(we see the output) so in that case as we can't input -4 and -1 so we wont be able to include the output of these two values?

You can't include something that does not exist. Stop thinking about pink unicorns, unruly gnomes, and the resultant of functions outside their domains. As for how to show the range, please see Sir Michael's post.

 

[Deleted member 4993]

...Stop thinking about pink unicorns, unruly gnomes, ....

Can we think about pink elephants .... white unicorns ..... singing gnomes....???

 

[JeffM]

Can we think about pink elephants .... white unicorns ..... singing gnomes....???

Only in a bar

 

[hamza4best]

It is true that there is a hole at (-4, -1/3), but also you must realize that the limit of this function as x goes to positive or negative infinity is 0 so the range is actually:

(-∞, -1/3) U (-1/3, 0) U (0, ∞)

What I know about finding range is that we look at three things

1. are we getting a +ive answer
2. are we getting a -ve answer
3. Can we get the function equal to 0(zero)

Since we get the first 2 conditions except for -1/3 as there is a hole at -4 so it won't be included
In you suggested range you didn't include 0 as we can't get the function equal to zero yes?

 

[JeffM]

What I know about finding range is that we look at three things

1. are we getting a +ive answer
2. are we getting a -ve answer
3. Can we get the function equal to 0(zero)

Since we get the first 2 conditions except for -1/3 as there is a hole at -4 so it won't be included
In you suggested range you didn't include 0 as we can't get the function equal to zero yes?

Hamza

Sir Michael may have further thoughts, but I do not like the way you are thinking about range. Determining a function's range is not about determining the sign of the function. It is about determining what, within the function's domain, are the limits on the function. Finding where the sign of the function changes may be very helpful to that determination, but it just a step in getting to the end result.

The domain of the function \(\displaystyle f(x) = \dfrac{x + 4}{x^2 + 5x + 4}\) has three sub-domains.

In each sub-domain \(\displaystyle f(x) = \dfrac{1}{x + 1}.\)

\(\displaystyle x < - 4 \implies x + 1 < - 3 \implies - (x + 1) > 3 > 0 \implies 0 < - \dfrac{1}{x + 1} < \dfrac{1}{3} \implies - \dfrac{1}{3} < \dfrac{1}{x + 1} < 0 \implies - \dfrac{1}{3} < f(x) < 0.\)

Furthermore, \(\displaystyle \displaystyle \lim_{x \rightarrow -\infty}f(x) = 0.\)

So the range corresponding to one subdomain is \(\displaystyle \left(- \dfrac{1}{3},\ 0\right).\)

Now do a similar analysis for the two other sub-domains.