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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 $h(h(\sqrt{2}))$.

 Aug 27, 2017
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\(\[h(x) = \left\{ \begin{array}{cl} \lfloor 4x \rfloor & \text{if } x \le \pi, \\ 3-x & \text{if }\pi < x \le 5.2, \\ x^2& \text{if }5.2< x. \end{array}\right.\]\)

 

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

 

 

 

cool cool cool

 Aug 28, 2017
edited by CPhill  Aug 28, 2017
edited by CPhill  Aug 28, 2017

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