+1242 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 \(f\left(\frac{\pi}{3}\right) + f(\sqrt{45}) + f(8^{2/3}).\)
+4609 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=\(\boxed{12}\).
