Question

use the intermediate value theorem to show that there is a positive number whose 5th power is exactly 1 more than itself.

Answer 1

So it's asking you to show that there is at least one number x, such that,

\[\Large x^5 = x+1\]

Answer 2

do you see what to do from here?

Answer 3

not exactly

Answer 4

would I show that it is continuous somewhere?

Answer 5

I could show that it is continuous at a positive number

Answer 6

are you allowed to use a graphing calculator?

Answer 7

not to show the solution

Answer 8

well you can use it to help find one, right?

Answer 9

it should be possible through intermediate value theorem

Answer 10

yes it should be

Answer 11

what I would do is use a graphing calculator to set up the proper interval

Answer 12

so you can get the right endpoints to test

Answer 13

the number lies between 1 and 2, but not sure where

Answer 14

you don't need to find the actual number

you just need to show it exists

Answer 15

could I set up the equations as y=x^5 and y=x+1?

Answer 16

through the use of the intermediate value theorem

Answer 17

no I would do it like this

x^5 = x+1

x^5 - x - 1 = 0

Answer 18

then use the intermediate value theorem to show there is at least one root between x = 1 and x = 2 for x^5 - x - 1 = 0

Answer 19

how might you do that?

Answer 20

let f(x) = x^5 - x - 1

what is f(1) equal to?

Answer 21

oh, I get it, crossing the axis means that one number existed between those two

Answer 22

f(1) = -1

Answer 23

f(2) = 29

Answer 24

one root had to have occurred at that interval, so...can you concisely define why that makes this true?

Answer 25

we have a sign change from - to + on the interval [1,2]

so somewhere in between, there is at least one root for x^5 - x - 1 = 0

Answer 26

the best way to think of it, at least for me, is to do so visually

Answer 27

if this function is continuous (which it is), then going from a negative y region to a positive y region must mean we cross the x axis at some point in between

so we might have something like this maybe

Answer 28

or maybe like this