Question

Please explain step by step.
Let f(x)= x/x^2+1 use the first derivative test to determine which critical numbers if any give relative extrema determine the intervals on which the function is increasing or decreasing by evaluating f'(x) at appropriate test values. Based on the sign of the derivative changes between intervals determine where the function has relative extrema .

Answer 1

Did you get the \(f'(x)\) ?

Answer 2

Yes

Answer 3

There are three cases when critical numbers occur.

\(\sf\color{red}{■ ~~~Case ~1}\)
If any closed-x-interval boundaries are given then,
these closed boundaries are critical numbers.

\(\sf\color{red}{■ ~~~Case ~2}\)
Any x-values that satisfy \(f'(x)=0\) are critical numbers.
(Well, if they are part of function's domain)

\(\sf\color{red}{■ ~~~Case ~3}\)
Any x-values at which \(f'(x)\) is undefined.
Provided that at these x-values the \(f(x)\) is defined.

Answer 4

For the derivative, [the typical thing here is to] use the "Quotient Rule"

Answer 5

What was your derivative, can you show me?

Answer 6

1/(x^2 +1) - 2x^2/(x^2+1)^2

Answer 7

But that can be simplified

Answer 8

that is rather likely, i mean the way you wrote, if you did,

\(\color{#000000 }{ \displaystyle \frac{d }{dx} \left[x(x^2+1)^{-1}\right]=(x)'(x^2+1)^{-1}+x\left((x^2+1)^{-1}\right)' }\)

\(\color{#000000 }{ \displaystyle \frac{d }{dx} \left[x(x^2+1)^{-1}\right]=1\cdot (x^2+1)^{-1}+x\left(-(x^2+1)^{-2}\times 2x\right) }\)

\(\color{#000000 }{ \displaystyle \frac{d }{dx} \left[x(x^2+1)^{-1}\right]=\frac{1}{x^2+1}+\frac{-2x^2}{(x^2+1)^2} }\)

Good job

Answer 9

is this how you did it?
(through product rule, with negative exponent)

I like that better than quotient, btw.

Answer 10

In any case, the derivative is right so let's move on.

Answer 11

Ok

Answer 12

\(\color{#000000 }{ \displaystyle f'(x)=\frac{1}{x^2+1}-\frac{ 2x^2 }{(x^2+1)^2} }\)

but, I think we would need to combine anyway

\(\color{#000000 }{ \displaystyle f'(x)=\frac{x^2+1}{(x^2+1)^2}-\frac{ 2x^2 }{(x^2+1)^2} }\)

\(\color{#000000 }{ \displaystyle f'(x)=\frac{x^2+1- 2x^2 }{(x^2+1)^2} }\)

that is simpler looking...

\(\color{#000000 }{ \displaystyle f'(x)=\frac{1-x^2 }{(x^2+1)^2} }\)

Answer 13

You are NOT given any closed-interval boundaries, so case 1 (for critical numbers) is excluded. Case 2 is also excluded, since \(x^2+1\) is defined for all real x.

Answer 14

So, only remains case 3.

Answer 15

\(\color{#000000 }{ \displaystyle 0=\frac{1-x^2 }{(x^2+1)^2} }\)

please solve for x.

Answer 16

One moment

Answer 17

it will be quick if you multiply both sides times \((x^2+1)^2\)

Answer 18

What.

Answer 19

\(\color{#000000 }{ \displaystyle 0=\frac{1-x^2 }{(x^2+1)^2} }\)

What I mean is,

\(\color{#000000 }{ \displaystyle 0~\color{red}{\times (x^2+1)^2}=\frac{1-x^2 }{(x^2+1)^2}~\color{red}{\times (x^2+1)^2} }\)

Answer 20

Isn't it just one?

Answer 21

For the left side you know that anything times 0 is equal to .....
For the right side, \((x^2+1)^2\) will cancel out.

Answer 22

x=1, is a solution, but not the only.

Answer 23

Ok how are their other solutions?

Answer 24

yes, there is another solution besides x=1.

Answer 25

\(\color{#000000 }{ \displaystyle 0=\frac{1-x^2 }{(x^2+1)^2} }\)

you should be capable of solving this.
(You got the derivative. This is so much simpler)

Answer 26

Negative 1?

Answer 27

yes:)

Answer 28

I was out because I was in the hospital when we were taught how to do these. I have a very one track mind.

Answer 29

Yay!

Answer 30

Thank you :)

Answer 31

\(\color{#000000 }{ \displaystyle 0=\frac{1-x^2 }{(x^2+1)^2} }\)

\(\color{#000000 }{ \displaystyle 0~\color{red}{\times (x^2+1)^2}=\frac{1-x^2 }{(x^2+1)^2}~\color{red}{\times (x^2+1)^2} }\)

\(\color{#000000 }{ \displaystyle 0=\frac{1-x^2 }{\cancel{(x^2+1)^2}}~\color{}{\times \cancel{(x^2+1)^2}} }\)

\(\color{#000000 }{ \displaystyle 0=1-x^2 }\)
\(\color{#000000 }{ \displaystyle x^2=1 }\)
(Will do it precisely, instead of making the \(\pm\) pop from nowhere)
I will take the square root of both sides,
\(\color{#000000 }{ \displaystyle \sqrt{x^2}=\sqrt{1} }\)
\(\color{#000000 }{ \displaystyle |x|=1 }\)
\(\color{#000000 }{ \displaystyle x=\pm 1 }\)

Answer 32

(That is the definition of absolute value \(|x|=\sqrt{x^2}\) (for real numbers))

Answer 33

So, your critical numbers are?

Answer 34

The absolute value of 1?

Answer 35

Or 1 and -1

Answer 36

yes, the critical numbers are;\
x=1
x=-1

Answer 37

And now, plug these numbers into the function and tell me what you get for each one of them.

Answer 38

1/2 and -1/2

Answer 39

Yes, very good.

Answer 40

In fact,

f(1)=1/2
is absolute maximum
(max of y=1/2 that occurs at x=1)

f(-1)=-1/2
is absolute minimumn
(min of y=-1.2 that occurs at x=-1)

(if I find a need, we will refer to this statement later to show it)

Answer 41

When the slope is negative the function is decreasing.
When the slope is positive the function is increasing.

Am I right?

Answer 42

I believe so XD

Answer 43

Yes.

Answer 44

The derivative is the [instantaneous] slope.

(This is by defintion, and if you want, I can precisely explain why this is so after we are done with this problem)

Answer 45

So, equivalently correct statements would be,

When the \(\color{#000000 }{ \sf derivative }\) is negative the function is decreasing.
When the \(\color{#000000 }{ \sf derivative }\) is positive the function is increasing.

Answer 46

(If you ever lose me, or if I am going too fat/too slow then let me know)

Answer 47

No, you're good

Answer 48

Ok, and we actually have the derivative expression/function.
So, we are able to figure when the slope/derivative is negative and when is positive.

(you know that positive means greater than 0, and negative means less than zero)

Answer 49

If you recall, I have simplified the derivative to be
(if yo don't recall, then simplifying this is fairly easy)

\(\color{#000000 }{ \displaystyle f'(x)=\frac{1-x^2 }{(x^2+1)^2} }\)

Answer 50

So for negative slopes

\(\color{#000000 }{ \displaystyle 0>\frac{1-x^2 }{(x^2+1)^2} }\)



and for positive slopes

\(\color{#000000 }{ \displaystyle 0<\frac{1-x^2 }{(x^2+1)^2} }\)

Answer 51

now, you need to solve and tell me where the slope/derivative is negative and where [the derivative] is positive.

Answer 52

So when x is equal to or less than 1 it's negative? And when x is equal to or greater than 2 it's positive?

Answer 53

well, no...

Answer 54

Negative slope,

\(\color{#000000 }{ \displaystyle 0>\frac{1-x^2 }{(x^2+1)^2} }\)

\(\color{#000000 }{ \displaystyle 0>1-x^2 }\)

\(\color{#000000 }{ \displaystyle x^2>1 }\)

\(\color{#000000 }{ \displaystyle \sqrt{x^2}>\sqrt{1} }\)

Answer 55

keep going from there please.

Answer 56

I honestly don't know what I'm supposed to do there

Answer 57

simplify that and solve for x.

Answer 58

I'm barley passing calculus and I barely even passed the past two years of math.

Answer 59

Then, I guess you have to review the previous maths untill you can answer any of these questions. You can't move on like this.

Answer 60

It's my last year of high school. I get through this and I can go back to algebra in collage

Answer 61

but, how can you do calculus without knowing algebra?

Answer 62

(btw, I have recently seen a user with a similar problem, and I think I motivated this user to review algebra)

Answer 63

I'd like to pass on the whole motivation thing. I had awful teachers until I changed schools and now my teachers are just trying to keep me afloat. I'm an A+ student.... But in math I'm more like a D student

Answer 64

It's hard enough without your judgement

Answer 65

Just look the common course outline for algebra and try to nail all of the things possible.

Answer 66

have a good look!

Answer 67

thank you for your time.

Answer 68

Anytime.