Answer
\( f(-0.5) \approx 20.56 \)
Solution
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.
Reviewed and approved by the UpStudy tutoring team
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