Let's solve the problem in a step-by-step manner.
### Step 1: Evaluate [tex]\( h(\sqrt{2}) \)[/tex]
First, we need to determine which case of the function [tex]\( h \)[/tex] we will use for [tex]\( \sqrt{2} \)[/tex].
[tex]\[
\sqrt{2} \approx 1.414
\][/tex]
Given that [tex]\( \sqrt{2} \leq \pi \)[/tex], we use the first case of the function:
[tex]\[
h(x) = \lfloor 4x \rfloor \quad \text{if} \quad x \leq \pi
\][/tex]
Substituting [tex]\( x = \sqrt{2} \)[/tex]:
[tex]\[
h(\sqrt{2}) = \lfloor 4 \cdot \sqrt{2} \rfloor
\][/tex]
Calculate [tex]\( 4 \cdot \sqrt{2} \)[/tex]:
[tex]\[
4 \times 1.414 \approx 5.656
\][/tex]
Applying the floor function:
[tex]\[
\lfloor 5.656 \rfloor = 5
\][/tex]
Thus, we have:
[tex]\[
h(\sqrt{2}) = 5
\][/tex]
### Step 2: Evaluate [tex]\( h(h(\sqrt{2})) \)[/tex], i.e., [tex]\( h(5) \)[/tex]
Now, we need to evaluate [tex]\( h(5) \)[/tex]. We determine which case of the function [tex]\( h \)[/tex] applies for [tex]\( x = 5 \)[/tex].
[tex]\[
\pi < 5 \leq 5.2
\][/tex]
Given that [tex]\( 5 \)[/tex] falls in the interval [tex]\( \pi < x \leq 5.2 \)[/tex], we use the second case of the function:
[tex]\[
h(x) = 3 - x \quad \text{if} \quad \pi < x \leq 5.2
\][/tex]
Substituting [tex]\( x = 5 \)[/tex]:
[tex]\[
h(5) = 3 - 5 = -2
\][/tex]
### Conclusion
We evaluated [tex]\( h(\sqrt{2}) \)[/tex] to be [tex]\( 5 \)[/tex], and then we evaluated [tex]\( h(5) \)[/tex] to be [tex]\( -2 \)[/tex].
Therefore, the final result is:
[tex]\[
h(h(\sqrt{2})) = -2
\][/tex]
