The function $h(x)$ is defined as: \[h(x) = \left\{ \begin{array}{cl} \lfloor 4x \rfloor & \text{if } x \le \pi, \\ 3-x & \text{if }\pi < x \le 5.2, \\ x^2& \text{if }5.2< x. \end{array}\right.\] Find $h(h(\sqrt{2}))$.
\(\[h(x) = \left\{ \begin{array}{cl} \lfloor 4x \rfloor & \text{if } x \le \pi, \\ 3-x & \text{if }\pi < x \le 5.2, \\ x^2& \text{if }5.2< x. \end{array}\right.\]\)
Find \($h(h(\sqrt{2}))$\)
We want to first evaluate h (√2)....note that √2 < pi ....so we want to use the first function
h (√2) = floor [ 4(√2) ] = floor [ ≈ 5.657 ] ....and we want the greatest integer < or = to 5.657
This is 5
Now ....we want to evaluate h(5).....this falls between pi and 5.2, so we want to use the second function
So h(5) = 3 - 5 = -2
So.....to recap
h ( h(√2) ) = h (5) = -2