Question

Help with intermediate value theorem pls

Answer 1

The Intermediate Value Theorem guarantees the existence of at least one real zero of the polynomial function f(x) = -x^3 − 4x + 1 on the following interval:

Answer 2

[-2, 0]

[1, 2]

[3, 6]

[-1, 1]

Answer 3

and the IVT states?

Answer 4

For a polynomial function f(x), if a value x = a where f(a) is positive, and another value x = b where f(b) is negative are found, it can then be concluded that the function has at least one real zero between those two values.

Answer 5

logically speaking, if we start walking along the side of the road and start to daydream ... then find ourselves on the other side of the road ... we must have corssed over the road at some point along the way

if the end points of our interval produce difference signs, then we would have had to corss over 0 at some point inbetween

Answer 6

plug in endpoints and see which switch signs

Answer 7

Wouldn't they all switch signs though><

Answer 8

test them out ....

Answer 9

f(-2)=-2^3-4(-2)+1
f(-2)=-8+8+1
f(-2)=1

Answer 10

6 more to go :) the wolf tells me the zero is about x=1/4

Answer 11

hokay

Answer 12

f(-2) = +1, f(0) = +1 no change in sign so no zero in between

Answer 13

f(0)=0-0+1
f(0)=1
so the first one is not true

Answer 14

correct

Answer 15

same signs doesnt say that there was NOT a zero, but rather that we simply cant tell for sure what happened along the interval

Answer 16

f(1)=-1^3-4(1)+1
f(1)=-1-4 +1
f(1)=-5+1
f(1)=-4

Answer 17

f(2)=-2^3 -4(2) +1
f(2)=-8-8+1
f(2)=1

Answer 18

soh this is false also

Answer 19

f(3)=-3^3-4(3)+1
f(3)=-27-12+1
f(3)=-38

Answer 20

f(1)=-(1)^3-4(1)+1
f(1)=-1-4 +1
f(1)=-5+1
f(1)=-4

f(2)= -(2)^3 -4(2) +1
f(2)= -8-8+1
f(2)= -16+1
f(2)= -15

which is why its false

Answer 21

i failed s: oops pesky - signs

Answer 22

f(6)=-6^3 -4(6)+1
f(6)=-215-24+1
f(6)=-238

Answer 23

soh this is false also

Answer 24

3 options down, only 1 left :)

Answer 25

f(-1)=1^3 - 4(-1) +1
f(-1)=1+4+1
f(-1)=6

f(1)=-1^3 -4(1)+1
f(1)=-1-4+1
f(1)=-4

Answer 26

just as a little side bar

we determined f(0) = 1 and f(1) = -4 beforehand; so if we had been vigilant then we would have seen that a 0 is between f(0) and f(1)

Answer 27

but we needed to 'chk' them all outright just to determine which one fit the IVT specifically

Answer 28

i guess it would make sense why the answer would be d if f(0)=1 and f(1)=-4

Answer 29

yep :)

Answer 30

aha thank you for helping :D this makes so much more sense now with evaluating it :p

Answer 31

good luck :)