Exercises: Terms with Square Roots

Only positive or zero expressions are allowed under the root. So $x-36$ must be positive or zero. $x-36$ becomes zero for $x=36$.

• When we make $x$ smaller (so $x<36$), then $x-36$ gets negative (which is not good)

• When we make $x$ larger (so $x>36$), then $x-36$ becomes positive (which is what we want).

So $x\ge 36$ is the required condition for obtaining a positive or zero number under the root. Further, $x$ is a real number, so

Only positive or zero expressions are allowed under the root. So $36+x^2$ must not be negative. Now, $x^2$ is never negative and if we add $+36$, it only becomes "more positive". So any $x$ is allowed in the domain of definition, as long as it is a real number:

Only positive or zero expressions are allowed under the root. So $x+36$ must not be negative. Further, we cannot divide by zero, which excludes the case $\sqrt{x+36}=0$, which happens if $x+36$ becomes zero. So only $x+36>0$ is allowed. This is the case whenever $x>-36$. Further, we know that $x$ is a real number.

So the domain of definition of $\frac{1}{\sqrt{\mathrm{x}+36}}$ is characterized by

Only positive or zero expressions are allowed under the root. So $x^2-36$ must not be negative. That means, $x^2$ must be larger or equal $36$.

• For $x=6$ and $x=-6$, you exactly obtain $x^2=36$. This is a "good case"

• If $x$ is positive, then making $x$ larger (so $x>6$) will make $x^2$ larger than $36$, which is also a "good case".

• Analogously, for negative $x$, making $x$ even more negative (so $x<-6$) will also result in $x^2>36$, which is good for our purposes.

• However, if -$6<x<6$, so $x\in (-6,6)$, then $x^2$ becomes smaller than 36, which is not good.. So we have to exclude this case.

Further, we know that $x$ is a real number. So the domain of definition is given by