Question

HELP ME!!!!!! For which value of n does the intermediate value theorem guarantee that f(x)= x^3+ 2x + 1 has a zero in the interval (n/2, (n+1)/2)

Answer 1

what does the intermediate value theorem say?

Answer 2

a continuous function does not skip vaues

Answer 3

Nope, that isn't what it says exactly, though that is somewhat of a corollary I suppose.

Answer 4

i agree

Answer 5

clearly im clueless thats what my calculus teacher told me

Answer 6

It says that if \(f(x)\) is continuous on the closed interval \([a,b]\), then there exists a \(c\) where \(f(a) \leq f(c) \leq f(b)\).

Answer 7

Now, if we're trying to prove a root exists, we'd want \(f(a) \leq 0\) and \(f(b) \geq 0\)

Answer 8

Then we'd have a root between \([a,b]\)

Answer 9

i get that there has to be a positive and a negative value i just dont know how to find it in this case given that the intervals have variables

Answer 10

You need to solve for \(n\).

Answer 11

so find f(n/2) and f(n+1/2)?

Answer 12

\[
\frac{n^3}{2^3}+\frac {2n}{2}+1\leq 0
\]

Answer 13

That is the equation \(f(n/2)\leq 0\).
The other one is \(f((n+1)/2) \geq 0\).

Answer 14

\[f(\frac{ n }{ 2})\le1\]

Answer 15

?

Answer 16

im lost again

Answer 17

\(\color{blue}{\text{Originally Posted by}}\) @wio
\[
\frac{n^3}{2^3}+\frac {2n}{2}+1\leq 0
\]
\(\color{blue}{\text{End of Quote}}\)
Find solutions for \(n\).

Answer 18

\[
n^3+8n+8\leq 0
\]

Answer 19

Actually, the easiest thing to do might be simply to find the roots of the actual polynomial first.

Answer 20

Find roots of \[
x^3+ 2x + 1
\]

Answer 21

Then you know: \[
\frac n2 \leq r \leq \frac{n+1}2
\]

Answer 22

Or rather: \[
n\leq 2r\leq n+1
\]

Answer 23

the root of \[x^3+ 2x+1 \] is \[\pm1\]

Answer 24

\(-1\) is a root. \(1\) isn't.

Answer 25

So you could do: \[
n\leq 2(-1)\leq n+1\implies n\leq -2 \leq n+1
\]You could let \(n=-2.5\)

Answer 26

And test that out and see if it works.

Answer 27

it didnt work would i just keep plugging in random numbers for n until it works?

Answer 28

Does \(n\) have ot be a natural number?

Answer 29

idk my homework is online and it tells me if the answer is wrong and i plugged in n=-2.5 and it said incorrect

Answer 30

Oh wait!

Answer 31

\(-1\) wasn't a root. It was \(-\frac 12\)

Answer 32

how did u figure that?

Answer 33

So: \[
n\leq -1\leq n+1
\]

Answer 34

are the numbers you're choosing random!

Answer 35

?

Answer 36

oh because its n/2

Answer 37

No, I already explained my logic.

Answer 38

Yes. that is why I multiply the root by \(2\)

Answer 39

ok so n=-1 thank you so much :)

Answer 40

so this problem is much easier than it seems