Question

Intermediate Value Theorem Question

Answer 1

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.

\[\sqrt[3]{x} = 1 - x, (0, 1)\]

Answer 2

So... when I plugged in for 0 and 1, I got 1 both times. Which means there is not an intersection, right? But the question sounds like they want me to prove the intersection, so I dunno

Answer 3

\[\large\rm \sqrt[3]{x}=1-x\]Subtracting the root to the other side,
and multiplying by a -1 gives us,\[\large\rm 0=\sqrt[3]{x}+x-1\]So there is some function... when you try to find it's root it looks like that ^
So think of the function like this,\[\large\rm f(x)=\sqrt[3]{x}+x-1\]And the problem might make a little more sense.

Answer 4

Yeah, that does make more sense. Thank you!

What about this one?
\[e^x = 3 - 2x, (0, 1)\]

Answer 5

I mean, yes, you have the right idea...
The problem is essentially telling you to prove that there is an intersection.
But the word "root" was to give you a hint to approach it this way.

Answer 6

Are you talking about the first Q or the 2nd one? o.O

Answer 7

Same approach for the next one, ya? :)

Answer 8

Both I suppose.

Answer 9

I think I have a handle on the first one, use the IVT etc etc.
The second one... e scares me D:
\[0 = e^x + 2x - 3\]

Answer 10

\[\large\rm f(x)=0=e^x+2x-3\]We're trying to prove that some function has a root.
So we compare f(0) and f(1),
(psst, one will be negative, the other positive),
and then by continuity we can apply IVT, ya?

Answer 11

Should I have the powers of e^x memorized? :O

Answer 12

e^0 would obviously be 1... e^1 would be e...

Answer 13

You should have e^0 and e^1 memorized... yes -_-

Answer 14

:D
What number does e stand for?

Answer 15

It's like... 2.71828 or something.
It's larger than -1 though right?
(which should be showing up somewhere in your calculations).

Answer 16

Hmm okay. Should I memorize the value of e?

Answer 17

Also... how would I know that the first equation was continuous?

Answer 18

I think knowing that it's around the same value as pi is probably good enough. Like when you're doing the calculations for this second problem, you should end up with something like this,\[\large\rm f(0)=-2\]\[\large\rm f(1)=e-1\]Since f(x) is a continuous function on \(\large\rm [0,1]\) and \(\large\rm f(0)<0<f(1)\),
the Intermediate Value Theorem guarantees there exists a \(\large\rm 0<c<1\) such that \(\large\rm f(c)=0\).

I hope I worded that correctly...

Answer 19

I guess you have to remember a little bit about `odd roots`.

Answer 20

`even roots` are the ones that cause problems.

Answer 21

Why is that? because they can produce two solutions?