Intermediate Value Theorem

hsiwin3

New member
Joined
Sep 10, 2009
Messages
8
Use IVT to find the interval(s) of length 1/8 containing a solution of the given equation. (Select all that apply.)
f (x) = x[sup:1knarxmy]3[/sup:1knarxmy] + 2x + 1/4

[-1, 0]
[-1/4, 0]
[-1/8, 0]
[0, 1]
[7/8, 1]
[3/4, 1]

I honestly have no idea where to start. My professor didn't explain at all and there are no example problems like this in the book.
 
I'm confused as well. The "equation" you were given is actually just a naming of the function. Its always true trivially. Second there are only two intervals of length 1/8 given. The correct answer to this question, as written, loos like the 3rd and 5th choices. You aren't asked to set f(x)=0 or anything?
 


We need to understand the Intermediate Value Theorem (IVT), before working any exercise based on it.

Given an interval [a, b], the IVT tells us that a zero exists between a and b if the signs of f(a) and f(b) are different.

As Loren notes, only two of the given intervals have length 1/8.

The fifth choice does not give opposite signs on f(x), so there is no zero for f between 7/8 and 1.

Check the third interval. In other words, compare the signs of f(-1/8) and f(0). 8-)

 
Top