# Using Rolle's Theorem find numbers

I am suppose to use Rolle's Theorem and then find all numbers c that satisfy the conclusion of the theorem.

$f(x) = x^4 +4x^2 + 1 [-3,3]$

Polynomials are always going to satisfy the theorem.

The derivative is

$4x^3 + 8x$ and the only number that could possibly make that zero would be zero so the answer is 0.

What did I get wrong? I failed my calc test and this was one of the one's I got wrong.

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You have, for $a = -3, b = 3$, $f(a) = f(b)$. By the theorem, there exists $c \in (-3, 3)$ such that $f \ '(c) = 0$. Solving $$4x^3 + 8x = 0$$ gives $x = 0$ as the only real solution... am I missing something too?? [Note - polynomials will always satisfy the continuous and differentiable conditions] –  The Chaz 2.0 Apr 17 '12 at 1:07
In general, you should ask your instructor what went wrong before other people. There are many reasons an answer might not be totally correct, and the instructor is the only one who can tell you exactly why you are wrong. That being said, Kannapan's answer would be my guess as to where you went wrong. –  user1306 Apr 17 '12 at 1:11
The instructor said that I didn't show enough work to prove that 4x^3 + 8x = 0, I am planning to petition the result I got on the test so I just wanted some clarification. –  user138246 Apr 17 '12 at 1:14
Perhaps they were looking for you to explain that $4x^2+8>0$ for all $x$? –  user1306 Apr 17 '12 at 1:20

Polynomials are always going to satisfy theorem.

No. They satisy the hypothesis about continuity and differentiability in the closed and open intervals respectively while you need to check if the polynomials agree at the end point.

Here they will agree on the end points because $x^2$ and $x^4$ are even functions and the interval is symmetric about $0$. (Do you see why?)

And, you are in need of $c$ that satisfies $f'(c)=0$. You're right but I would be surprised if the instructor does not want you to prove your claim:

\begin{align} f'(c)&=0 \;\;\mbox{and $c \in (-3,3)$} \tag{1}\\4c^3+8c&=0\\4c(c^2+2)&=0 \end{align}

Since $c^2+2 \ge 2$ for $c \in \Bbb R$, the equation $c^2+2=0$ does not have roots in $\Bbb R$ and hence $c=0$ is the only value that satisfies $(1)$.

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Jordan, if you wrote "the only number that could possibly make $4x^3 + 8x = 0$ is zero" on my test, I would have a hard time giving you full credit. It's true, but you need to show some reasoning (or at least factoring!). –  The Chaz 2.0 Apr 17 '12 at 1:18
@TheChaz Just to clarify, did I not convey the same sentiments about justifying the answers? –  user21436 Apr 17 '12 at 1:34
You sure did, but I wanted to underscore that point :) –  The Chaz 2.0 Apr 17 '12 at 1:35
I do not know what that large R means, or the E looking symbol. –  user138246 Apr 17 '12 at 1:43
@Jordan The $\Bbb R$ stands for the set of all real numbers. The $\in$ denotes membership in a set. We write $a \in A$ to mean that $a$ is an element of the set $A$. Concretely, $a \in \Bbb R$ means $a$ is a real number. –  user21436 Apr 17 '12 at 1:47