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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 :

Let's determine the values of the function [tex]\( h \)[/tex] at the given points [tex]\( x = 0 \)[/tex] and [tex]\( x = 4 \)[/tex].

First, we evaluate [tex]\( h(0) \)[/tex]. According to the piecewise function definition:
[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 second case:
[tex]\[ h(0) = 2(0)^2 - 3(0) + 10 = 10 \][/tex]

Next, we evaluate [tex]\( h(4) \)[/tex]. According to the piecewise function definition:
[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]\( x = 4 \)[/tex], we use the third case:
[tex]\[ h(4) = 2^4 = 16 \][/tex]

Therefore, the values of the function are:
[tex]\[ \begin{array}{l} h(0) = 10 \\ h(4) = 16 \end{array} \][/tex]
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