Question

< closed I guess? >
Two Questions -

(1) S̶h̶o̶w̶ ̶t̶h̶a̶t̶ ̶t̶h̶e̶ ̶e̶q̶u̶a̶t̶i̶o̶n̶ ̶s̶i̶n̶x̶-̶6̶x̶=̶0̶ ̶h̶a̶s̶ ̶e̶x̶a̶c̶t̶l̶y̶ ̶o̶n̶e̶ ̶r̶o̶o̶t̶ ̶u̶s̶i̶n̶g̶ ̶t̶h̶e̶ ̶M̶e̶a̶n̶ ̶V̶a̶l̶u̶e̶ ̶T̶h̶e̶o̶r̶e̶m̶.̶.̶.̶
I̶ ̶d̶o̶n̶'̶t̶ ̶k̶n̶o̶w̶ ̶h̶o̶w̶ ̶t̶o̶ ̶a̶p̶p̶l̶y̶ ̶t̶h̶e̶ ̶M̶V̶T̶?̶

(2) Find the following:
a. intervals of increase/decrease
b. local min/max values
c. intervals of concavity and inflection points

function: f(x)=ln(x^4+1)

Answer 1

LaTeX:
(1) \(\sin{(x)}-6x=0\)
(2) \(f(x)=\ln{(x^{4}+1})\)

Answer 2

The idea is to:

1) Note that the function is continuous and differentiable in the Domain where you care.
2) Find a value of x such that sin(x) - 6x > 0
3) Find a value of x such that sin(x) - 6x < 0
4) The Mean Value Theorem (MVT) tells us that there must be a value where sin(x) - 6x = 0.

However, there is no guarantee there is ONLY one such value. I'd check on the specific wording of the question.

Answer 3

Er... I don't understand what step/part 1 means, exactly. How would I be able to tell that it's differentiable on an interval...?

Or is that what steps 2 and 3 are explaining?

Answer 4

You tell me. Is it differentiable? Is ti continuous?

Answer 5

I don't understand the question...

Answer 6

f(x) = sin(x) - 6x

Is there a derivative on some interval?
Is there anywhere where it jumps around or is it well-behaved?

Answer 7

I understand continuity but not the derivative part*
Sorry.

Answer 8

How are you supposed to solve Problem #2? Are you studying the Derivative?

Answer 9

That problem is from a section titled "How Derivatives Affect the Shape of a Graph"

Answer 10

I get part (a) but the rest I am not sure if I need to apply certain concepts...

Answer 11

Okay, then you should have some idea what a derivative is.

Does f(x) = sin(x) - 6x have a derivative?

Answer 12

I think that it's \(f'(x)=\cos{(x)}-6\)?

Answer 13

There you go. Now, on to step #2.

Answer 14

The first problem is from "Mean Value Theorem"
The second problem is from "How Derivatives Affect the Shape of a Graph"

Answer 15

Okay

Answer 16

Are steps 2 and 3 variable? (any number that makes said equation true)*

Answer 17

Any value will do.

Answer 18

Alright, thank you. I will work on that... might take a while

Answer 19

Nah. Try x = +1 and x = -1.

Answer 20

Oh... alright lol

Just asking... is the MVT related to the IVT in any way?

Answer 21

The are similar, but not to be confused.

Answer 22

Okay, so I got \(f(1)\approx-5.157529...\) and \(f(-1)\approx5.15829...\)

Answer 23

ok, so you know the function f(x)=sin(x) - 6x , is continuous everywhere, and goes from a negative value to a positive value, so it must cross f(x)=0 somewhere At Least Once between -1 and 1.

From the intermediate value theorem, you can say there is At least one root inside the interval [-1,1]. That doesn't say that there cant be more than one though...

Answer 24

You can show if there is ONLY one root by assuming there are more and testing to see if it is true. Assume there are 2 roots, and apply the mean value theorem...

Answer 25

If you have 2 roots where f(x)=0, say x=a and x=b, then the mean value theorem will say that there must be some value between a and b where the derivative is zero...

\[\large f'(c)=\frac{ f(b)-f(a) }{ b-a }\]
If f(a) and f(b) are both roots then
\[f'(c)=\frac{ 0-0 }{ b-a }=0\]

The function is continuous, so it must turn around somewhere in order to get 2 roots...

Answer 26

Notice the derivative..

f '(x) = cos(x) - 6

For two roots to exist for f(x)=sin(x)-6x , the derivative must be zero at least once..

cos(x) - 6 = 0

That never happens, cos(x)=6, not true.
This says that there aren't 2 or more roots for the function.

Answer 27

Overall...
From Intermediate value theorem, shown AT Least one Root exists.
From Mean Value Theorem, shown impossible for 2 or more roots to exist.
From those two, you can say exactly one root Exists.

Answer 28

http://openstudy.com/study#/updates/58288f0ee4b038c47dd29987

Answer 29

@LunaCat please refrain from posting links to your own questions on other people's posts! Thank you ☻

Answer 30

Could be wrong, but I have the strong feeling that you're discussing the Intermediate Value Theorem here, NOT the Mean Value Theorem. Please double-check on this.

"Is f(x) differentiable?" is not a step, but rather a CONDITION. Your function must satisfy at least two conditions before you can even consider applying the IVT or the MVT.

Answer 31

@mathmale to whom are you referring? Sorry, I am confused.

Answer 32

(2) Find the following:
a. intervals of increase/decrease
b. local min/max values
c. intervals of concavity and inflection points

a. where f'(x) > 0
b. where f'(x) = 0
c. where f''(x) = 0 (inflection) and where f''(x) < or > 0

Answer 33

Okay, thank you @issimplcalcus ! :)

Answer 34

I am referring to you, kittiwitti1.

Answer 35

The question is in the section "Mean Value Theorem" @mathmale

Answer 36

The MVT and the IVT are usually discussed in the same section. The MVT is, in my opinion, the more important of the two.

It doesn't surprise me if you find a problem in this MVT section that asks you to show that there's only one root for the given fn. on the given interval. But the IVT is the relevant theorem here, NOT the MVT.

Answer 37

I would like to see the problem on the new post. It is hard to read a long comment to follow what is going on.

Answer 38

@Loser66 "Show that the equation \(\sin{(x)}-6x=0\) has exactly one root (using MVT)"

Answer 39

I'm not quite sure how exactly he wants it, excluding the fact that he needs it done using MVT... :(

Answer 40

Im dumb whoops

Answer 41

1. MVT to show a zero exists
2. show (dy/dx) is < 0 for all x

Answer 42

eh? O-o

Answer 43

1. MVT to show a zero exists
2. Show that f(x) is only decreasing by showing f'(x) is < 0 for all x

Answer 44

OHH, I see. o:

Answer 45

You need one value of x which makes f(x) >0
And f(x) monotonic decreases as what issimplcalcus says

Answer 46

Alright, thank you both! I will work with that in mind. :-)