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I am confused as to what my book intends for me to do here. It is asking me to let $f(x)= 1-x^{2/3}$ and show that $f(-1)=f(1)$ but there there is no number $c$ in $(-1,1)$ such that the derivative is equal to $0$. Also why does this not contradict Rolle's Theorem?

I am getting stuck on finding a way to make those two functions equal.

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    The derivatives are equal, but that does not make the functions equal just the tangent line at points 1 and -1 I believe.2011-10-07
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    Rolle's theorem requires derivability on $(-1,1)$. Did you check if it's the case?2011-10-07
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    $f(1) = 1 - 1^{2/3} = 1-1 = 0$ and $f(-1) = 1 - (-1)^{2/3} = 1-1^{1/3} = 0$. I can't see your problem here.2011-10-07
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    Yes, the fact that $f(1)=f(-1)$ is true. But is $f$ differentiable on $(-1,1)$?2011-10-07
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    I get 1 and 2 for -1 and 1.2011-10-07
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    $1-(1^{2/3})=0$ and $1-(-1^{2/3})=2$2011-10-07
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    @Jordan: The question is: does $f$ have a derivative defined **everywhere** on $(-1,1)$? What is the derivative of $f$? You can probably figure that one out easily enough. Now look at the formula you got. Does that formula make sense for *every* number between $-1$ and $1$?2011-10-07
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    @Jordan: No: $$(-1)^{2/3} = \sqrt[3]{(-1)^2} = \sqrt[3]{1}=1$$so $1-(-1)^{2/3} = 1-1 = 0$. Note that $(-1)^{2/3}$ is not the same thing as $-1^{2/3}$; the latter is $-(1^{2/3})$.2011-10-07
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    I get a derivative of $2/3 x ^{-1/3}$2011-10-07
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    I have $ 1-(-1^{2/3}) = 2$ and $1-(1^{2/3})=0$ Wolfram is confirming this as well.2011-10-07
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    @Jordan: Again, you are computing $$1 - ( - (1^{2/3}))$$instead of $$1 - ((-1)^{2/3}).$$Put the parenthesis in the **correct place**. $$(-1)^{2/3} = \sqrt[3]{(-1)^2} = \sqrt[3]{1} = 1.$$2011-10-07
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    I don't understand the appearance of the double parantheses infront of -1.2011-10-07
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    @Jordan: If you write $-1^a$, then because exponentiation must be done before you do products, Wolfram Alpha first computes $1^a$, then multiplies the answer by $-1$. So when you write $-1^{2/3}$, WolframAlpha first computes $1^{2/3}$, which is $1$, then multiplies by $-1$ and gets $-1$. **But this is not how you must evaluate this function at $-1$.** In this function, **first** you need to take $-1$, **then** raise it to the $2/3$rd power. So you need to do $(-1)^{2/3}$, not $-1^{2/3}$. **They are not the same thing**, just like $5\times 2+3$ is not the same as $5\times(2+3)$.2011-10-07
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    Okay so they are all 1-1=0 so they are equal.2011-10-07
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    It's not a "double parenthesis". Perhaps this way it will be clearer? You are computing $$1 - \Biggl( -\quad \bigl( 1^{2/3}\bigr)\quad\Biggr).$$ What you *should* be computing is $$1 - \Biggl(\quad \bigl(-1\bigr)^{2/3}\Biggr).$$And to compute $(-1)^{2/3}$, we do $\left(\ (-1)^2\ \right)^{1/3}$.2011-10-07
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    Sorry I forgot basic math again.2011-10-07
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    "[...] make those two functions equal." This line of text shows that there is something you have gravely misunderstood. The value of a function (from $\mathbb{R}$ to $\mathbb{R}$) at a point is not itself a function. I did not understand at all what you were talking about until I read through the comments and realised that what you were having problems with was verifying that $f(-1) = f(1)$.2011-10-07
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    What is R? How is it not a function? It is a function with a defined variable.2011-10-07
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    $\mathbb{R}$ is the set of all real numbers. $f(1)$ and $f(-1)$ are not functions, they are *numbers*: they are the **result** of evaluating a function, and as such they are numbers. Neither $f(1)$ nor $f(-1)$ are "functions". Just like $g(x) = x^2$ is a function, but $g(2)$ is not a function, it's a number (namely, $4$).2011-10-07

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First, you probably intended to write $f(x) = 1-(x^2)^{1/3}$, so that $f(-x) = f(x)$, $\forall x \in \mathbb{R}$.

You already computed the derivative, but notice that there is a point where the derivative is not defined. Indeed, computing the derivative using the chain rule: $$ f^\prime(x) = \left( \frac{1}{3} (x^2)^{-2/3} \right) \cdot ( 2 x) = \frac{2}{3} \frac{x}{(x^2)^{2/3}} $$ Notice how $f^\prime(x) < 0$ for $x <0 $ and $f^\prime(x) > 0$ for $x>0$. Consider what happens as $x$ approaches the origin from the left and from the right.

You should see now why $f(-1) = f(1) = 0$ does not violate Rolle's theorem.

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    Why are you writing the exponents like that?2011-10-07
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    @Jordan: By definition, $a^{p/q}$ with $p$ and $q$ inters is $\sqrt[q]{a^p}$; that is, $a^{p/q} = (a^p)^{1/q}$.2011-10-07
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    So $(x^2)^{-2/3}$ is equal to $2/3x^{-1/3}$?2011-10-07
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    @Jordan I did write it as $(x^2)^{1/3}$ to emphasize that given a number $x$, we first square it, and them raise to power of $1/3$. This way $f(x)$ is explicitly symmetric.2011-10-07
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    The difference between $x^{2/3}$ and $(x^2)^{1/3}$ show when you extend the function to the complex plane. The standard way to interpret $x^{2/3} = \exp( \frac{2}{3} \ln(x))$. So for $x=-1$, $(-1)^{2/3} = \exp( \frac{2}{3} \ln(-1)) = \exp( \frac{2}{3} i \pi ) = \cos( \frac{2 \pi}{3} ) + i \sin( \frac{2 \pi}{3} ) = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$.2011-10-07
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    What does exp mean?2011-10-07
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    @Sasha: I would advice against that detour; look at Jordan's posting history, and you'll see that is far more likely to confuse him thoroughly than to make anything clear.2011-10-07
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    I can't really imagine what happens to x as it approaches 0 but I am getting it gets close to or is 0 since both 1 and -1 are equal to zero.2011-10-07
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    Neither $1$ nor $-1$ are equal to $0$. $1$ is equal to $1$, $-1$ is equal to $-1$. You are probably trying to say that $f(1)$ and $f(-1)$ are both equal to zero (true but irrelevant); but that sloppiness is just one of the reasons why you are not doing well in this course. Mathematics does *not* forgive that kind of muddled language and muddled thinking. And the question you should be asking is "what happens to the **derivative** $f'(x)$ as $x$ approaches $0$?" The derivative, as Sasha has provided, is$$f'(x) = \frac{2x}{3(x^2)^{2/3}}=\frac{2}{3x^{1/3}}.$$ What happens as $x\to 0$?2011-10-07
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    The closer it gets to zero the larger it gets.2011-10-07
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    @Jordan What about the sign of $f^\prime(x)$ as $x$ approaches the origin2011-10-07
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    It is positive I think.2011-10-07
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    @Sasha sorry I am just not sure what I am suppose to be seeing. I am going to attempt to graph this.2011-10-08