To analyze the functions given by Clara—namely \(p(x) = 1.3x\), \(q(x) = 1.3x + 10\), and \(t(x) = 1.3^x\)—let's consider their growth rates.
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Linear functions \(p(x)\) and \(q(x)\):
- The function \(p(x) = 1.3x\) is a linear function representing a constant growth rate of 1.3.
- The function \(q(x) = 1.3x + 10\) has the same linear growth rate of 1.3, but it is always 10 units higher than \(p(x)\).
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Exponential function \(t(x)\):
- The function \(t(x) = 1.3^x\) is an exponential function, which starts off growing slower than both linear functions for small values of \(x\) but eventually grows much faster as \(x\) increases.
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Comparing growth rates:
- For small values of \(x\), \(t(x) = 1.3^x\) will initially be less than \(p(x) = 1.3x\) and \(q(x) = 1.3x + 10\) since \(1.3^x\) approaches 1 as \(x\) approaches 0, while both \(p(x)\) and \(q(x)\) start from values greater than 1.
- However, as \(x\) increases, \(t(x)\) will eventually surpass both \(p(x)\) and \(q(x)\).
Based on this analysis, the most accurate observation Clara could make is:
For small values of x, the growth rate of t(x)=1.3^x is less than the growth rate of p(x)=1.3x and the growth rate of q(x)=1.3x+10.
This statement reflects the behavior of the functions correctly, as \(t(x)\) is indeed slower than both \(p(x)\) and \(q(x)\) at the start, which aligns with the characteristics of exponential versus linear growth.
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