# Single variable calculus beginner - is this interval contained in the domain of the function?

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.

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

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 $2^a - a^3>0$

• "if a 0 then $-a^3$ > 0" this is not true as .02 < 0 but -.02 * -.02 * -.02 < 0 ? Aug 24, 2016 at 16:50
• @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. Aug 24, 2016 at 17:08

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 $(-\infty,0)\subset D$.

• how did you arrive at $2^x - x^3$ from $\sqrt{2^x - x^3}$ ? Aug 24, 2016 at 16:47
• @blue-sky The argument of a square root should be always $\geq 0$. Aug 24, 2016 at 16:52

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 $[0,\sqrt{2}].$

• how did you arrive at $2^x >= x^3$ from $\sqrt{2^x - x^3}$ ? Aug 24, 2016 at 17:30
• 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.) Aug 24, 2016 at 17:42