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Consider function [tex]$h$[/tex].
[tex]$
h(x)=\left\{\begin{array}{ll}
3x-4, & x\ \textless \ 0 \\
2x^2-3x+10, & 0 \leq x\ \textless \ 4 \\
2^x, & x \geq 4
\end{array}\right.
$[/tex]

What are the values of the function when [tex]$x=0$[/tex] and when [tex]$x=4$[/tex]?

[tex]$
\begin{array}{l}
h(0)=\square \\
h(4)=\square
\end{array}
$[/tex]


Sagot :

To determine the values of the piecewise function [tex]\( h(x) \)[/tex] for [tex]\( x = 0 \)[/tex] and [tex]\( x = 4 \)[/tex], we will evaluate the function as follows:

1. Evaluating [tex]\( h(0) \)[/tex]:

According to the given piecewise function:
[tex]\[ h(x) = \begin{cases} 3x - 4 & \text{if} \; x < 0 \\ 2x^2 - 3x + 10 & \text{if} \; 0 \leq x < 4 \\ 2^x & \text{if} \; x \geq 4 \end{cases} \][/tex]

For [tex]\( x = 0 \)[/tex], we use the piece [tex]\( 2x^2 - 3x + 10 \)[/tex] because [tex]\( 0 \leq x < 4 \)[/tex].
[tex]\[ h(0) = 2(0)^2 - 3(0) + 10 = 0 - 0 + 10 = 10 \][/tex]

So, [tex]\( h(0) = 10 \)[/tex].

2. Evaluating [tex]\( h(4) \)[/tex]:

For [tex]\( x = 4 \)[/tex], we use the piece [tex]\( 2^x \)[/tex] as [tex]\( x \geq 4 \)[/tex].
[tex]\[ h(4) = 2^4 = 16 \][/tex]

So, [tex]\( h(4) = 16 \)[/tex].

Putting these together, we have:
[tex]\[ \begin{array}{l} h(0) = 10 \\ h(4) = 16 \end{array} \][/tex]

Thus, the values of the function are:
[tex]\[ h(0) = 10 \\ h(4) = 16 \][/tex]
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