Question

Assume that abs[f '(x)] ≤ M for a ≤ x ≤ b. One can show by the Mean Value Theorem that abs[f(b) − f(a)] ≤ M (b − a), and thus f(a) − M(b − a) ≤ f(b) ≤ f(a) + M(b − a). Using this fact, determine a lower bound and an upper bound for sqrt101 (Hint: Let f (x) = sqrt(x), a = 100, and b = 101, and then find bounds for f(101) = sqrt(101).
First, find a reasonable value for M: M = ______

Answer 1

No idea what M even is. We didn't talk a bout this in class and its not in the textbook

Answer 2

@wio please help <.>

Answer 3

So like, I would think that M could be any almost any positive number for the equation to be true, but I don't know what is considered "reasonable"

Answer 4

\[
|f'(x)|<M
\]Using mean value theorem on this gives us \[
\left|\frac{f(b)-f(a)}{b-a}\right|<M
\]

Answer 5

And \[
\frac{|f(b)-f(a)|}{|b-a|}
\]Now since \(a\leq b\) then \(0\leq b-a\) so \(b-a\) is zero or positive meaning \(|b-a|=b-a\)

Answer 6

\[
\frac{|f(b)-f(a)|}{b-a}<M
\]And \[
|f(b)-f(a)|<M(b-a)
\]

Answer 7

\(M\) is just some number, a bound for the derivative

Answer 8

So sqrt 101- sqrt 100?

And then just plug that in?

Answer 9

I thought that too satellite, but I plugged in 1, and it said that that was wrong.

Answer 10

We have two equations: \[
f(b)-f(a)<M(b-a)
\]And \[
f(a)-f(b)<M(b-a)
\]

Answer 11

no it is not 1
you are asked for a bound for \(f'(x)=\frac{1}{2\sqrt{x}}\)

Answer 12

since \(x>100\) i guess you can use \(\frac{1}{2\sqrt{100}}=\frac{1}{20}\)

Answer 13

that is just a guess, mind you, i would not bet on it

Answer 14

\[
f'(x)
\]Is a monotonic decreasing function, so it is highest at the beginning of any interval.

Answer 15

So letting \(M=f'(100)\) is fine here.

Answer 16

And \(b-a=1\)

Answer 17

Okay, I see why 1/20 satisfies the f(b)-f(a) < M part, but why can it not be greater than that?

Answer 18

No, \[
|f(b)-f(a)|< M(b-a)
\]

Answer 19

So |sqrt101-sqrt100| ends up being about .5, and (b-a) is 1 so you can throw htat out

So ~.05 <M

Why can M not be any number greater than .05?

Answer 20

There is no bounds on M

Answer 21

M can be one billion.

Answer 22

The online thing that I'm using said that a value of 1 for M was wrong.

Answer 23

It only has a lower bounds: it can't be lower than the derivative.

Answer 24

Well, the reason is that they want you to be more precise.

Answer 25

M can be as big as you want, but you want M to be a low as possible in order to tighten the bounds around \(f(b)\) as much as possible.

Answer 26

So it's not that my answer was necesarily wrong, it was just that it wasn't good enough?

Do I just need to guess at what they think is a small enough value?

Answer 27

Well, there is a smallest value M can be.

Answer 28

The smallest value M can be and still satisfy the requirements, is the the absolute extrema in the interval.

Answer 29

They want the smallest M, most likely.

Answer 30

And that would be sqrt101 -sqrt 100 right?

They don't allow me to enter in a squareroot, and that is an irrational number right?

Answer 31

Nope, wrong.

They want M to be the largest that \(f'(x)\) gets in the inverval \([a,b]\).

Answer 32

Since \(f'(x)\) is an always decreasing function, then it's maximum would be at the beginning of the interval, in other words \(f'(a)\).

Answer 33

Which evaluates to \(1/20\).

Answer 34

Oooohh

Answer 35

I see

Answer 36

That makes a lot of sense. Sorry I was being so dumb with this.

Answer 37

yeah. That was the right answer. Thank you so much dude