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Sagot :
To identify the function type that could represent the data in the table, we need to carefully analyze the relationship between [tex]\( x \)[/tex] and [tex]\( f(x) \)[/tex].
Given data:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline 1 & 4 \\ 4 & 8 \\ 9 & 12 \\ 16 & 16 \\ 25 & 20 \\ \hline \end{array} \][/tex]
First, observe the pattern in the [tex]\( x \)[/tex]-values: [tex]\( 1, 4, 9, 16, 25 \)[/tex]. These values are perfect squares (i.e., [tex]\( x \)[/tex] is increasing quadratically as [tex]\( x = 1^2, 2^2, 3^2, 4^2, 5^2 \)[/tex]).
Next, look at the [tex]\( f(x) \)[/tex]-values: [tex]\( 4, 8, 12, 16, 20 \)[/tex]. These values are increasing linearly.
We suspect that [tex]\( f(x) \)[/tex] may be related to the square root of [tex]\( x \)[/tex] since while [tex]\( x \)[/tex] is increasing quadratically, [tex]\( f(x) \)[/tex] increases linearly. To explore this, let's denote [tex]\( y = \sqrt{x} \)[/tex], and rewrite the table:
[tex]\[ \begin{array}{|c|c|} \hline y = \sqrt{x} & f(x) \\ \hline 1 & 4 \\ 2 & 8 \\ 3 & 12 \\ 4 & 16 \\ 5 & 20 \\ \hline \end{array} \][/tex]
We can see that there might be a linear relationship between [tex]\( y \)[/tex] (which is [tex]\( \sqrt{x} \)[/tex]) and [tex]\( f(x) \)[/tex]. Let's hypothesize that [tex]\( f(x) \)[/tex] is of the form:
[tex]\[ f(x) = a \cdot \sqrt{x} + b \][/tex]
To find the constants [tex]\( a \)[/tex] and [tex]\( b \)[/tex], we use the pairs [tex]\((y, f(x))\)[/tex]:
[tex]\[ \begin{array}{c|c} y & f(x) \\ \hline 1 & 4 \\ 2 & 8 \\ 3 & 12 \\ 4 & 16 \\ 5 & 20 \\ \end{array} \][/tex]
We solve for [tex]\( a \)[/tex] and [tex]\( b \)[/tex] using these pairs:
For [tex]\( y = 1 \)[/tex], [tex]\( f(1) = 4 \)[/tex]:
[tex]\[ 4 = a \cdot 1 + b \quad \Rightarrow \quad a + b = 4 \quad \text{(Equation 1)} \][/tex]
For [tex]\( y = 2 \)[/tex], [tex]\( f(4) = 8 \)[/tex]:
[tex]\[ 8 = a \cdot 2 + b \quad \Rightarrow \quad 2a + b = 8 \quad \text{(Equation 2)} \][/tex]
To solve this system of linear equations, subtract Equation 1 from Equation 2:
[tex]\[ (2a + b) - (a + b) = 8 - 4 \][/tex]
[tex]\[ a = 4 \][/tex]
Substitute [tex]\( a \)[/tex] back into Equation 1:
[tex]\[ a + b = 4 \][/tex]
[tex]\[ 4 + b = 4 \][/tex]
[tex]\[ b = 0 \][/tex]
So, the function can be written as:
[tex]\[ f(x) = 4 \sqrt{x} + 0 \][/tex]
Thus, the linear function type that represents the data in the table is:
[tex]\[ f(x) = 4 \sqrt{x} \][/tex]
This function shows that [tex]\( f(x) \)[/tex] increases linearly with the square root of [tex]\( x \)[/tex].
Given data:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline 1 & 4 \\ 4 & 8 \\ 9 & 12 \\ 16 & 16 \\ 25 & 20 \\ \hline \end{array} \][/tex]
First, observe the pattern in the [tex]\( x \)[/tex]-values: [tex]\( 1, 4, 9, 16, 25 \)[/tex]. These values are perfect squares (i.e., [tex]\( x \)[/tex] is increasing quadratically as [tex]\( x = 1^2, 2^2, 3^2, 4^2, 5^2 \)[/tex]).
Next, look at the [tex]\( f(x) \)[/tex]-values: [tex]\( 4, 8, 12, 16, 20 \)[/tex]. These values are increasing linearly.
We suspect that [tex]\( f(x) \)[/tex] may be related to the square root of [tex]\( x \)[/tex] since while [tex]\( x \)[/tex] is increasing quadratically, [tex]\( f(x) \)[/tex] increases linearly. To explore this, let's denote [tex]\( y = \sqrt{x} \)[/tex], and rewrite the table:
[tex]\[ \begin{array}{|c|c|} \hline y = \sqrt{x} & f(x) \\ \hline 1 & 4 \\ 2 & 8 \\ 3 & 12 \\ 4 & 16 \\ 5 & 20 \\ \hline \end{array} \][/tex]
We can see that there might be a linear relationship between [tex]\( y \)[/tex] (which is [tex]\( \sqrt{x} \)[/tex]) and [tex]\( f(x) \)[/tex]. Let's hypothesize that [tex]\( f(x) \)[/tex] is of the form:
[tex]\[ f(x) = a \cdot \sqrt{x} + b \][/tex]
To find the constants [tex]\( a \)[/tex] and [tex]\( b \)[/tex], we use the pairs [tex]\((y, f(x))\)[/tex]:
[tex]\[ \begin{array}{c|c} y & f(x) \\ \hline 1 & 4 \\ 2 & 8 \\ 3 & 12 \\ 4 & 16 \\ 5 & 20 \\ \end{array} \][/tex]
We solve for [tex]\( a \)[/tex] and [tex]\( b \)[/tex] using these pairs:
For [tex]\( y = 1 \)[/tex], [tex]\( f(1) = 4 \)[/tex]:
[tex]\[ 4 = a \cdot 1 + b \quad \Rightarrow \quad a + b = 4 \quad \text{(Equation 1)} \][/tex]
For [tex]\( y = 2 \)[/tex], [tex]\( f(4) = 8 \)[/tex]:
[tex]\[ 8 = a \cdot 2 + b \quad \Rightarrow \quad 2a + b = 8 \quad \text{(Equation 2)} \][/tex]
To solve this system of linear equations, subtract Equation 1 from Equation 2:
[tex]\[ (2a + b) - (a + b) = 8 - 4 \][/tex]
[tex]\[ a = 4 \][/tex]
Substitute [tex]\( a \)[/tex] back into Equation 1:
[tex]\[ a + b = 4 \][/tex]
[tex]\[ 4 + b = 4 \][/tex]
[tex]\[ b = 0 \][/tex]
So, the function can be written as:
[tex]\[ f(x) = 4 \sqrt{x} + 0 \][/tex]
Thus, the linear function type that represents the data in the table is:
[tex]\[ f(x) = 4 \sqrt{x} \][/tex]
This function shows that [tex]\( f(x) \)[/tex] increases linearly with the square root of [tex]\( x \)[/tex].
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