# Help of Intermediate Value Theorem

• Jun 13th 2006, 02:22 PM
c_323_h
Help of Intermediate Value Theorem
Use the Intermediate Value Theorem to show that $\displaystyle x^5-x^2+1=0$has a solution.

Given:
$\displaystyle f(-1)=-1$ and $\displaystyle f(0)=1$

I understand what the IVT is and how to find a approximate solution using trial and error, but what exactly does it mean by "show?" Are they asking for a proof? I understand the concept of IVT but I can't translate my potential way of solving this into math symbols and such

Thanks
• Jun 13th 2006, 02:32 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
Use the Intermediate Value Theorem to show that $\displaystyle x^5-x^2+1=0$has a solution.

Given:
$\displaystyle f(-1)=-1$ and $\displaystyle f(0)=1$

The function,
$\displaystyle f(x)=x^5-x^2+1$ is countinous on $\displaystyle [-1,0]$
By IVT,
For any $\displaystyle -1=f(-1)\leq k\leq f(0)=1$ there exists an $\displaystyle 1 \leq x \leq 0$ such as, $\displaystyle f(x)=k$
Let $\displaystyle k=0$ and the proof is finished.
• Jun 13th 2006, 02:55 PM
c_323_h
Quote:

Originally Posted by ThePerfectHacker
The function,
$\displaystyle f(x)=x^5-x^2+1$ is countinous on $\displaystyle [-1,0]$
By IVT,
For any $\displaystyle -1=f(-1)\leq k\leq f(0)=1$ there exists an $\displaystyle 1 \leq x \leq 0$ such as, $\displaystyle f(x)=k$
Let $\displaystyle k=0$ and the proof is finished.

This is what I tried. Let $\displaystyle f(x)=x^5-x^2+1=0$ . Then, by the IVT, there exists a number $\displaystyle c$ such that
$\displaystyle f(-1)<f(c)<f(0)$. Since the function is continuous over the interval $\displaystyle (-1,1)$ there exists a number $\displaystyle N$ such that $\displaystyle -1<N<1$ and $\displaystyle f(c)=N$

I don't know if that proves anything though. I really don't have a direction when I try to prove things. I just copied the definition of IVT out of the book.
• Jun 13th 2006, 04:48 PM
ThePerfectHacker
Let me try this again.

If $\displaystyle f(x)$ is a real function countinous on closed interval $\displaystyle [a,b]$, and let $\displaystyle k$ be any number between $\displaystyle f(a)$ and $\displaystyle f(b)$ then there must exist an $\displaystyle x$ on $\displaystyle [a,b]$ such as, $\displaystyle f(x)=k$.

Root location theorem: If $\displaystyle f(x)$ is countinous on $\displaystyle [a,b]$ and $\displaystyle f(a)f(b)<0$ (in other words opposite signs) then there is a solution to the equation $\displaystyle f(x)=0$ on $\displaystyle [a,b]$

Proof: By the condition we have that $\displaystyle f(a)$ and $\displaystyle f(b)$ have opposite signs. Thus, zero is a number between them. Thus, by the intermediate value theorem, $\displaystyle f(x)=0$ for some $\displaystyle x$ on $\displaystyle [a,b]$
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1)Given $\displaystyle f(x)=x^5-x^2+x+1$ on $\displaystyle [-1,0]$
3)And, $\displaystyle f(-1)=-1$ and $\displaystyle f(0)=1$
4)And, $\displaystyle f(-1)f(0)<0$
6)Thus there exists a solution to $\displaystyle f(x)=0$.