# Thread: 2 problems with continous functions

1. ## 2 problems with continous functions

1) Show that 2^x = 3x for some x element (0,1)

2) show that any polynomial of odd degree has at least one real root

2. Originally Posted by luckyc1423
1) Show that 2^x = 3x for some x element (0,1)
Here's 1)

Consider the function f(x) = 2^x – 3x

f(0) = 1
f(1) = -1

Therefore, by the intermediate value theorem, there exist some c between 0 and 1, such that:
f(c) = 0
that is, for x = c, 0<c<1
2^x – 3x = 0
=> 2^x = 3x

3. Originally Posted by luckyc1423
2) show that any polynomial of odd degree has at least one real root
This is the same idea as Jhevon's solution to the first problem.

Given y(x) = ax^n + ... + b, where a is not 0 and n is odd, note that
lim[y(x), x --> -(infinity)] --> (-/+)(infinity) <-- Choice of sign depends on the sign of the coefficient a.
lim[y(x), x--> (infinity)] --> (+/-)(infinity)

Since the polynomial function either goes from -(infinity) to (infinity) or from (infinity) to -(infinity) we know it has to cross the x-axis at least once.

-Dan

4. Originally Posted by luckyc1423
2) show that any polynomial of odd degree has at least one real root
Consider the odd degree polynomial, P(x) = ax^n + bx^(n-1) + cx^(n-2) + ..., where n is odd.

Assume a>0

Now P(k) = ak^n + bk^(n-1) + ck^(n-2) + ...
=> P(k) = ak^n(1 + [bk^(n-1)]/ak^n + [ck^(n-2)]/ak^n + ...)
=> P(k) = ak^n(1 + b/ak + c/ak^2 + ...)
=> lim{k--> -inf}P(k) = lim{k--> -inf}ak^n(1 + b/ak + c/ak^2 + ...) = lim{k--> -inf}aK^n

since n is odd,
=> lim{k--> -inf}ak^n = -inf
So there is some k for which P(k) < 0

Now P(l) = al^n + bl^(n-1) + cl^(n-2) + ...
=> P(l) = al^n(1 + [bl^(n-1)]/al^n + [cl^(n-2)]/al^n + ...)
=> P(l) = al^n(1 + b/al + c/al^2 + ...)
=> lim{l--> inf}P(l) = lim{l--> inf}al^n(1 + b/al + c/al^2 + ...) = lim{l--> inf}al^n

since n is odd,
=> lim{l--> inf}al^n = inf
So there is some l for which P(l) > 0

Now, the opposte happens for a < 0, that is, P(k)>0 and P(l)<0, either way, 0 is between P(k) and P(l). Thus by the intermediate value theorem, there is some c between k and l such that P(c) = 0. So for x = c, P(x) has a real root

EDIT: Ah man, Dan beat me to it. I went a lot slower for the second one than he did though, I guess you can use his way