Given the problem, we are tasked with finding the exponential function $f(x) = ab^x$ based on the provided points $f(3.5) = 8$ and $f(7) = 33$. We then need to determine $f(6.5)$ to the nearest hundredth.

Step 1: Set up the system of equations

Using the general exponential form $f(x) = ab^x$, we can set up two equations based on the information given:

1. $f(3.5) = 8$, so: $ab^{3.5} = 8$

2. $f(7) = 33$, so: $ab^7 = 33$

Step 2: Solve for $a$ and $b$

To eliminate $a$, divide the second equation by the first:

$\frac{ab^7}{ab^{3.5}} = \frac{33}{8}$ This simplifies to: $b^{7 - 3.5} = \frac{33}{8}$ $b^{3.5} = \frac{33}{8} \approx 4.125$

Next, solve for $b$ by taking the logarithm of both sides: $b = \left( 4.125 \right)^{1/3.5}$ Using a calculator: $b \approx 1.426$

Step 3: Solve for $a$

Substitute the value of $b$ back into one of the original equations, such as $ab^{3.5} = 8$: $a(1.426)^{3.5} = 8$ Calculate $(1.426)^{3.5}$: $1.426^{3.5} \approx 2.851$ Now solve for $a$: $a \times 2.851 = 8$ $a \approx \frac{8}{2.851} \approx 2.806$

Step 4: Find $f(6.5)$

Now that we have the values of $a$ and $b$, we can find $f(6.5)$: $f(6.5) = 2.806 \times (1.426)^{6.5}$ First, calculate $1.426^{6.5}$: $1.426^{6.5} \approx 6.162$ Now, multiply by $a$: $f(6.5) \approx 2.806 \times 6.162 \approx 17.29$

Final Answer:

The value of $f(6.5)$ is approximately 17.29.

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Do you want further details or have any questions?

Here are 5 relative questions for further exploration:

1. How do exponential functions differ from linear functions?
2. What are the properties of the base $b$ in an exponential function?
3. How does logarithmic scaling relate to solving exponential equations?
4. Could the exponential model be used in real-world growth scenarios?
5. What is the significance of base $e$ in natural exponential functions?

Tip: Always check your calculations by plugging values back into the original equations to verify consistency!