1. Feb 21, 2014

### TitoSmooth

I did not understand the Intermediate Value Theorem explained today in class. The professor did not even go to his office hours :(.

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

#51 in Stewart's Calculus 7th ed. (section 1.8)

x^4 + x - 3=0

What I know so far. the equation or function has to be continuous throughout the given interval.

ie (0,2) so f(0)<c<f(1)

we plug in the x's into the equation and get the Y values then we compare them.

my problem is i do not understand the computation.

Can someone please give me a step by step approach the problem? Thanks.

2. Feb 21, 2014

### Mentallic

So you're trying to find where $x^4+x-3=0$.

Trying x=0 gives us $0^4+0-3=-3$ and x=2 gives $2^4+2-3=16+2-3=15$

So at x=0, the y value is -3, and at x=2 the y value is 15. This means that somewhere in the interval (0,2) the function must have cut across the x-axis (y=0) because the function is continuous and needs to go from a negative value at x=0 to a positive value at x=2.

If you try x=1 you'll get $1^4+1-3=-1$ so then the interval that the root is located can be shortened to (1,2). You can then test, say, 1.5 and see whether that's a positive of negative value. If it's negative, then the interval is (1.5,2) but if it's positive, then the interval is (1,1.5).

3. Feb 21, 2014

### TitoSmooth

so if i got it right.

we evaluate the function. 1<c<2

f(1)= - value
f(2)= + value

so now, we take the average of 1&2. which gives us x=1.5 (dont ask me why proffesor told us to do it this way in lecture)

now, we find f(1.5)

f(1.5)= + value. we compare f(1) & f(1.5.)-----> 1<c<1.5

take the average of 1 &1.5. gives us x=1.25

f(1.25)= + value. then we compare f(1.25) & f(1.5)

yet f(1.25) &f(1.5) are both positive. The theorem if I understand means it is between a negative and positive value.

so my answer is actually 1<c<1.5 ?

4. Feb 21, 2014

### Mentallic

Why would you compare f(1.25) and f(1.5)? If f(1.25) is positive, then the root c must lie between 1 and 1.25, so you then calculate the function value at the average of those x values which is f(1.125), and continue on with that process. We are always comparing two y values that are opposite in sign, because the function must go from positive to negative (or vice versa) hence it must have a root in that interval.

The reason we find the average of the values is because it's an easy algorithm to follow. Realistically, you could make educated guesses such as knowing that since f(1) = -1 and f(2) = 15 then the root is very likely to be closer to x=1 than x=2.

5. Feb 21, 2014

### TitoSmooth

the reason I compared f(1.25) and f(1.5) was to see if the signs were opposite of each other. which was not the case. So I reverted to (1,1.25) in the last line of my previous post. Thanks

6. Feb 21, 2014

### Ray Vickson

The reason your prof. asked you to do it this way is that he/she is introducing you to one of the standard equation-solving methods called "Bisecting Search". It is absolutely reliable, even in the worst case (unlike some faster methods) and yet its convergence rate is reasonably attractive. Every time you evaluate f(x) at a new point you reduce the interval of uncertainty by 50%. There are numerous much faster methods, but almost all of them can fail on some problems.