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Let\( f(n) = \left\{ \begin{array}{cl} \lfloor n \rfloor & \text{ if }n \geq 4, \\ \lceil n \rceil & \text{ if } n < 4. \end{array} \right.\)

 

Find \(f\left(\frac{\pi}{3}\right) + f(\sqrt{45}) + f(8^{2/3}).\)

 Aug 21, 2018
 #1
avatar+3994 
+3

We calculate, first,  pi/3=3.14/3 which is around 1.04, so ceiling function of that equals 2.

 

Next, the square root of 45 is greater than 6(6^2=36), and less than 7(7^2=49). Since it is greater than 4, the answer is 6.

 

Lastly, 8^2/3=(2^3)^2/3=2^2=4, which equals 4, Floor of that equals 4.

 

So, the answer is 2+6+4=8+4=\(\boxed{12}\).

 Aug 21, 2018
edited by tertre  Aug 21, 2018
 #2
avatar+98130 
+2

Nice, tertre....!!!!!!

 

 

 

cool cool cool

CPhill  Aug 21, 2018
 #3
avatar+3994 
+3

Thanks, CPhill!

tertre  Aug 21, 2018
 #4
avatar+973 
+1

Thanks! your right!

Lightning  Aug 21, 2018

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