If \( f(x) \) is an exponential function of the form of \( y=a b^{x} \) where \( f(-1.5)=12 \) and \( f(2)=79 \), then find the value of \( f(-0.5) \), to the nearest hundredth.

Given that \( f(x) \) is an exponential function of the form \( y = ab^x \), where \( f(-1.5) = 12 \) and \( f(2) = 79 \), we need to find the value of \( f(-0.5) \) to the nearest hundredth. To find the value of \( f(-0.5) \), we first need to determine the values of \( a \) and \( b \) using the given information. We have two equations: 1. \( f(-1.5) = 12 \) 2. \( f(2) = 79 \) Substitute the values of \( x \) into the exponential function: 1. \( 12 = ab^{-1.5} \) 2. \( 79 = ab^2 \) Now, we can solve these equations to find the values of \( a \) and \( b \). Let's solve the system of equations to find the values of \( a \) and \( b \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}12=ab^{-1.5}\\79=ab^{2}\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}12=ab^{-\frac{3}{2}}\\79=ab^{2}\end{array}\right.\) - step2: Solve the equation: \(\left\{ \begin{array}{l}a=12b^{\frac{3}{2}}\\79=ab^{2}\end{array}\right.\) - step3: Substitute the value of \(a:\) \(79=12b^{\frac{3}{2}}\times b^{2}\) - step4: Multiply the terms: \(79=12b^{\frac{7}{2}}\) - step5: Swap the sides: \(12b^{\frac{7}{2}}=79\) - step6: Divide both sides: \(\frac{12b^{\frac{7}{2}}}{12}=\frac{79}{12}\) - step7: Divide the numbers: \(b^{\frac{7}{2}}=\frac{79}{12}\) - step8: Raise both sides to the \(\frac{2}{7}\)-th power\(:\) \(\left(b^{\frac{7}{2}}\right)^{\frac{2}{7}}=\left(\frac{79}{12}\right)^{\frac{2}{7}}\) - step9: Evaluate the power: \(b=\frac{\sqrt[7]{6241\times 12^{5}}}{12}\) - step10: Substitute the value of \(b:\) \(a=12\left(\frac{\sqrt[7]{6241\times 12^{5}}}{12}\right)^{\frac{3}{2}}\) - step11: Simplify: \(a=\frac{\sqrt[14]{2187\left(6241\times 12^{5}\right)^{3}}}{6}\) - step12: Calculate: \(a=\frac{\sqrt[14]{2187\times 6241^{3}\times 12^{15}}}{6}\) - step13: Calculate: \(\left\{ \begin{array}{l}a=\frac{\sqrt[14]{2187\times 6241^{3}\times 12^{15}}}{6}\\b=\frac{\sqrt[7]{6241\times 12^{5}}}{12}\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}a=\frac{\sqrt[14]{2187\times 6241^{3}\times 12^{15}}}{6}\\b=\frac{\sqrt[7]{6241\times 12^{5}}}{12}\end{array}\right.\) - step15: Rewrite: \(\left(a,b\right) = \left(\frac{\sqrt[14]{2187\times 6241^{3}\times 12^{15}}}{6},\frac{\sqrt[7]{6241\times 12^{5}}}{12}\right)\) The solution to the system of equations is \( a = \frac{\sqrt[14]{2187 \times 6241^3 \times 12^{15}}}{6} \) and \( b = \frac{\sqrt[7]{6241 \times 12^5}}{12} \). Now, we can substitute these values back into the exponential function to find \( f(-0.5) \). Substitute \( a \) and \( b \) into the exponential function: \( f(-0.5) = \frac{\sqrt[14]{2187 \times 6241^3 \times 12^{15}}}{6} \times \left(\frac{\sqrt[7]{6241 \times 12^5}}{12}\right)^{-0.5} \) Let's calculate the value of \( f(-0.5) \) to the nearest hundredth. Calculate the value by following steps: - step0: Calculate: \(\frac{\sqrt[14]{2187\times 6241^{3}\times 12^{15}}}{6}\left(\frac{\sqrt[7]{6241\times 12^{5}}}{12}\right)^{-0.5}\) - step1: Simplify the root: \(\frac{12\sqrt[14]{26244\times 79^{6}}}{6}\left(\frac{\sqrt[7]{6241\times 12^{5}}}{12}\right)^{-0.5}\) - step2: Divide the terms: \(2\sqrt[14]{26244\times 79^{6}}\times \left(\frac{\sqrt[7]{6241\times 12^{5}}}{12}\right)^{-0.5}\) - step3: Convert the expressions: \(2\sqrt[14]{26244\times 79^{6}}\times \left(\frac{\sqrt[7]{6241\times 12^{5}}}{12}\right)^{-\frac{1}{2}}\) - step4: Multiply the terms: \(\frac{\sqrt[14]{2187\times 79^{12}\times 12^{9}}}{237}\sqrt[14]{26244\times 79^{6}}\) - step5: Multiply: \(\frac{\sqrt[14]{2187\times 79^{12}\times 12^{9}}\times \sqrt[14]{26244\times 79^{6}}}{237}\) - step6: Multiply: \(\frac{\sqrt[14]{57395628\times 79^{18}\times 12^{9}}}{237}\) The value of \( f(-0.5) \) to the nearest hundredth is approximately 20.56.