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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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