Reachin' the ceiling

Algebra Level 1

If $f(x) = \lceil x^{2}\rceil$ , what is the value of $f\left(\sqrt {\sqrt {5}} \right)$ ?

Notation : $\lceil \cdot \rceil$ denotes the ceiling function .

The answer is 3.

1 solution

Margaret Zheng
May 14, 2016

Relevant wiki: Ceiling Function

In this particular case, the value of x is $\sqrt {\sqrt {5}}$ . It is easy to understand that $x^{2}$ = $\sqrt {5}$ .

Now, because the function want us to find $\lceil \sqrt {5}\rceil$ , we need to find the smallest integer that is greater than $\sqrt {5}$ . because $2^{2} < 5 < 3^{2}$ , it is clear that this integer would be $\boxed {3}$ , which is our answer.

You have a mistake, its x=sqrt(5) and not x^2=sqrt(5)

אביחי וקנין - 4 years, 11 months ago

Well... did you see that there are 2 sqrt signs?

Margaret Zheng - 4 years, 11 months ago

You have made a mistake, it's x=sqrt(5) and not x^2=sqrt(5)

Am Kemplin - 16 hours ago

Reachin' the ceiling

The answer is 3.

1 solution

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