Function Value Combination
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}) $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
We begin by finding $ f\left(\frac{\pi}{3}\right) $. Since we know that $ \pi \approx 3.14 $, $ \frac{\pi}{3} $ must be slightly greater than $ 1 $, so $ f\left( \frac{\pi}{3} \right)= \left\lceil \frac{\pi}{3} \right\rceil = 2 $. To find $ f(\sqrt{45}) $, we realize that $ \sqrt{36} < \sqrt{45} < \sqrt{49} $, so $ 6 < \sqrt{45} < 7 $. Therefore, since $ \sqrt{45} \geq 4 $, we have that $ f(\sqrt{45}) = \lfloor \sqrt{45} \rfloor = 6 $. Finally, we consider $ f(8^{2/3}) $. We know that $ 8^{2/3}= (\sqrt[3]{8})^2 = 2^2 = 4 $, so $ f(8^{2/3})= \lfloor 8^{2/3} \rfloor = 4 $. Thus, we have that $ f\left(\frac{\pi}{3}\right) + f(\sqrt{45}) + f(8^{2/3}) = 2 + 6 + 4 = \boxed{12} $.
Freeflows