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I'm having a bit of trouble with something most everyone might find trivial, and I feel rather silly asking, but here it goes. The premise is as follows:

"Use the Intermediate Value Theorem to prove that the number $\sqrt[3]{20}$ exists and has a value greater than 2 but less than 3."

As part of the solution to the problem, the author establishes that $f(x)=x^3, a=2, b=3, f(a)=f(2)=8$, and $f(b)=f(3)=27$

My question is: As you can see in the premise, the function itself, $f(x)=x^3$ (and $f(2)$ or $f(3)$ for that matter) was not given, was there enough information in the problem itself to allow me to figure that out and I just wasn't able to do so as the solution is implying, or was the premise incomplete or wrongly worded? I'm inclined to think the problem is on my end, since this is what I consider to be a great book and an author I highly respect.

Thank you for your attention.

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    $\begingroup$ The información was enough, the function is "naturally" given by the index of the root. If the problem had been "show that for a given $n$, $\sqrt[n]{20}$ exists and satisfies a given inequality , you could use $f(x)=x^n$ and look for the values of n-powers nearest to 20. $\endgroup$ – Ángela Flores May 7 '17 at 19:38
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I'm sorry to say that, but, yes, there was enough information given.

The symbol $\sqrt[3]{20}$ denotes by definition a number $w$ with the the property that $e^3=20$. Therefore, it is natural to find $e$ by looking for a solution of the equation $$x^3=20.$$ In the context of the Intermediate Value Theorem (as mentioned in the problem statement), it is quite evident, that we are supposed to view the left hand side of this equation as a function (that we easily recognize as being continuous, just as needed for the theorem), and the constant on right hand as the function value. Recalling the statement of the IVT, the obvious task then is to find real numbers $a,b$ with $f(a)<20<f(b)$. And we do not even have to look far for these as they are literally given in the prblem statement as $2$ and $3$.

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    $\begingroup$ I think you should have stuck with $w$ instead of $e$, surely? Especially as $e^3 \doteq 20.086$, and one might get confused. $\endgroup$ – Brian Tung May 7 '17 at 19:46
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I think you have asked a very good question: Essentially, is a demonstration that assumes $2^3 = 8$ and $3^3 = 27$ a circular one? (At least, I think that's what you're asking. Perhaps not. As others have pointed out, the fact that $x^3$ is not given as the function is not an issue. It is up to you and your intuition to come up with that one.)

It may not be apparent to you at the moment, but the situation is that one can determine $2^3 = 8$ and $3^3 = 27$ with basic rules of arithmetic, which are rather a lower bar than the intermediate value theorem. So you may compute $2^3 = 2 \times 2 \times 2 = 4 \times 2 = 8$ in good conscience (and likewise with $3^3$), and use those values in your demonstration.

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Hint: Can we conclude that the function $$ f(x)=x^3-20 $$ has a root? Note that it is continuous and approaches positive infinity at infinity and negative infinity at negative infinity.

Equivalently and in the authors formulation, you can say $x^3=20$ at some point between 2 and 3 by noting that $x^3$ is continuous and is larger than 20 at 3 and smaller than 20 at 2.

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I would say the crucial point here that the author has not mentioned because he felt it too obvious is that cubing and cube root are inverse functions to each other.

If we to find a solution to $$ g(x) = y, $$ and $f\circ g = g \circ f = I$ then we have solution if and only if $$ x = f\circ g(x) = f(y). $$ So here we work with the inverse function which is a much nicer function.

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