If \( f(x) \) is an exponential function of the form of \( y=a b^{x} \) where \( f(3.5)=21 \) and \( f(8.5)=37 \), then find the value of \( f(14) \), to the nearest hundredth.
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To solve the problem, we first set up the equations based on the information given. The function is of the form \( f(x) = a b^x \). 1. From \( f(3.5) = 21 \): \[ ab^{3.5} = 21 \quad \text{(1)} \] 2. From \( f(8.5) = 37 \): \[ ab^{8.5} = 37 \quad \text{(2)} \] Next, we can divide equation (2) by equation (1) to eliminate \( a \): \[ \frac{ab^{8.5}}{ab^{3.5}} = \frac{37}{21} \] This simplifies to: \[ b^{8.5 - 3.5} = \frac{37}{21} \Rightarrow b^5 = \frac{37}{21} \] Now, let's find \( b \): \[ b = \left( \frac{37}{21} \right)^{\frac{1}{5}} \] Next, we substitute \( b \) back into equation (1) to find \( a \). We can rewrite equation (1) as: \[ a = \frac{21}{b^{3.5}} \] Substituting in our expression for \( b \): \[ a = \frac{21}{\left( \left(\frac{37}{21}\right)^{\frac{1}{5}} \right)^{3.5}} \] This becomes: \[ a = \frac{21}{\left( \frac{37}{21} \right)^{\frac{3.5}{5}}} = 21 \cdot \left( \frac{21}{37} \right)^{\frac{3.5}{5}} \] To find \( f(14) \): \[ f(14) = ab^{14} \] Using \( a \) and \( b \) we found earlier: \[ f(14) = \left( 21 \cdot \left( \frac{21}{37} \right)^{\frac{3.5}{5}} \right) \cdot \left( \frac{37}{21} \right)^{\frac{14}{5}} \] Calculating \( b^{14} \): \[ b^{14} = \left( \frac{37}{21} \right)^{\frac{14}{5}} \] Substituting this into our equation gives: \[ f(14) = 21 \cdot \left( \frac{21}{37} \right)^{\frac{3.5}{5}} \cdot \left( \frac{37}{21} \right)^{\frac{14}{5}} \] Now, we can find a simplified expression for \( f(14) \): \[ f(14) = 21 \cdot \frac{21^{\frac{14}{5}}}{37^{\frac{14}{5}}} \cdot \frac{37^{\frac{3.5}{5}}}{21^{\frac{3.5}{5}}} = 21^{1 + \frac{14}{5} - \frac{3.5}{5}} \cdot 37^{\frac{3.5}{5} - \frac{14}{5}} \] Calculating the new powers: \[ = 21^{(1 + \frac{14}{5} - \frac{3.5}{5})} \cdot 37^{(\frac{3.5}{5} - \frac{14}{5})} = 21^{\frac{5.0 + 14 - 3.5}{5}} \cdot 37^{\frac{3.5 - 14}{5}} \] At this stage, you would numerically compute the values of \( f(14) \) using a calculator for exact results. Upon performing these calculations, you will find that: \[ f(14) \approx 69.38 \] Thus, the final answer, rounded to the nearest hundredth, is: \[ \boxed{69.38} \]