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I'm taking a calculus course and a question is given :

Which of the following intervals are contained in the domain of the function :

$ \sqrt{2^x - x^3} $

1. $ [0 , \sqrt2] $

2. $ (-\infty , -\sqrt2) $


Here is how I'm approaching an answer.

  1. is a closed interval and so includes its endpoints on x.
  2. is an open interval and so does not include its endpoints.

$ [0 , \sqrt2] $ is two values for x and so I can substitute : $ \sqrt{2^0 - x^{\sqrt2}} $ ?

Is this a method to compute the answer ? : I compute the value of $ \sqrt{2^0 - x^{\sqrt2}} $ and if it is between 0 and $\sqrt2$ inclusive then the interval is contained within the domain.

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    $\begingroup$ Do you mean $\left(-\infty, -\sqrt{2}\right)$ for the second interval? $\endgroup$
    – Hrhm
    Aug 24, 2016 at 16:29
  • $\begingroup$ @Hrhm I meant the open interval, ive updated question thanks. $\endgroup$
    – blue-sky
    Aug 24, 2016 at 16:31
  • $\begingroup$ No problem, glad to help :) $\endgroup$
    – Hrhm
    Aug 24, 2016 at 16:31

3 Answers 3

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The only problem that can arise if you get a negative under the square root. Try plugging in $\sqrt{2}$ into your calculator for $2^x -x^3$ and if it is negative then that interval is bad.

Then try plugging some points from the other interval in as well. Do you ever get a negative under the square root?

Hint: if $a<0$ then $-a^3>0$ and so is $2^a - a^3>0$

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  • $\begingroup$ "if a 0 then $-a^3$ > 0" this is not true as .02 < 0 but -.02 * -.02 * -.02 < 0 ? $\endgroup$
    – blue-sky
    Aug 24, 2016 at 16:50
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    $\begingroup$ @blue-sky: $0.02$ is not less than $0$. If you mean $-0.02$, it is true that $(-0.02)^3$ is less than $0$, but note that you are subtracting that negative number from $2^{-0.02}$, so the result is positive. $\endgroup$
    – Brian Tung
    Aug 24, 2016 at 17:08
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The domain $D$ of your function is given by the set of $x$ such that $f(x)\geq 0$ where $f(x)=2^x-x^3$ is the argument of the square root.

We have that $\sqrt{2}\not \in D$ because $f(\sqrt{2})=2^{\sqrt{2}}-2^{3/2}<0$ ($3/2>\sqrt{2}$).

On the other hand, if $x<0$ then $2^x$ and $-x^3$ are positive. Therefore $f(x)>0$ and $(-\infty,0)\subset D$.

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  • $\begingroup$ how did you arrive at $2^x - x^3$ from $\sqrt{2^x - x^3}$ ? $\endgroup$
    – blue-sky
    Aug 24, 2016 at 16:47
  • $\begingroup$ @blue-sky The argument of a square root should be always $\geq 0$. $\endgroup$
    – Robert Z
    Aug 24, 2016 at 16:52
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You've got a lot going on here. First, in $\sqrt{2^x - x^3}$, whatever you plug in for one $x$ has to be plugged in all the $x$'s. So you can't plug $0$ in one of the $x$ and $\sqrt{2}$ for the other. Second, when you plug in, you need to plug into the $x$, in you changed the exponent $3$ to $\sqrt{2}$. You'd have two pluggings in: For $x=0$, you get $\sqrt{2^0 - 0^3} = 1$. For $x=\sqrt{2}$ you get $\sqrt{2^{\sqrt{2}} - \sqrt{2}^3} = \sqrt{-.16}$ which is not real.

That aside, to find the (natural) domain of a function, one normally starts with the reals and throws out points which break the function. E.g., divisions by 0, square roots of negatives, log's of non-positives, arcsin of things bigger than 1, etc. In this case, you need $2^x-x^3$ to be non-negative. If you graph both functions you can see when $2^x \geq x^3$ and when it's not. For $x$ to be in the domain, you must have $2^x \geq x^3.$ This doesn't happen for the interval $[0,\sqrt{2}].$

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  • $\begingroup$ how did you arrive at $2^x >= x^3$ from $\sqrt{2^x - x^3}$ ? $\endgroup$
    – blue-sky
    Aug 24, 2016 at 17:30
  • $\begingroup$ In most courses called "calculus", the study is restricted to real numbers. If you take the square root of a negative number, you no longer have a real number. So the stuff inside you're square root symbol needs to be at least zero. (Calculus courses involving complex numbers are usually called "Complex Variables" or some similar phrase.) $\endgroup$
    – B. Goddard
    Aug 24, 2016 at 17:42

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