Thread: Help with proof, thank youj.

Hey, I really apprecite this board. Thanks in advance everyone.
This is my question...
1. Let f (x) = x^7 +x-3. Find an integer n such that f (x) = 0 for some x between n and n + 1. How many real roots does f (x) have? Justify your answer.

Can someone explain this entire problem in proof format to compare it to my own. Thanks

Descartes' Rule of Signs says there is one real positive root. The FTA says there are 7 roots, so the rest are complex.

You can tell from the graph 'whereabouts' it lies. Therefore, you should be able to find n and n+1 from that.

Also, try Newton's method with an initial guess of 1. That'll give it to you and converges rather quickly.

The Intermediate Value Theorem may also be an course to take.

For that, try x=1 and x=2. Respectively, they give you -1 and 127. So, you know it's in between 1 and 2.

If you try it at x=1.1 you get 0.0487171. So, you see it's between 1 and 1.1. Keep whittling.

More accurately, the root is at

Originally Posted by galactus

Descartes' Rule of Signs says there is one real positive root. The FTA says there are 7 roots, so the rest are complex.

While your final answer is right the logic has a gap. You have to justify
the assertion that the rest are complex.

You can substitute -y for x to get -y^7-y-3 then the rule of signs says
there are no negative roots.

Or you can observe that if x is negative x^7+x-3 is negative and so there
are no negative roots.

As x=0 is obviously not a root we can now conclude that the remaining
roots are complex. (if x=0 were a root that would of course have left
an odd number of roots unaccounted for which is also not possible).

RonL

Yes, I realize that, Cap'N. I just got lazy as to listing all that.

On this problem, can someone help me finish it off with a set of logical conclusions explaining why it is between 1 and 2. And a little more help on how the answer x = 1.90633. Just basically helping me finish this proof off. Thank you