We are given that $f(x)$ is an exponential function of the form:

$f(x) = ab^x$

where $f(0.5) = 16$ and $f(7) = 38$. We need to find $f(1)$.

Step 1: Express the function using the given form

The general form of the exponential function is:

$f(x) = ab^x$

We can use the two known points $f(0.5) = 16$ and $f(7) = 38$ to form two equations.

1. At $x = 0.5$, we have: $f(0.5) = ab^{0.5} = 16$
2. At $x = 7$, we have: $f(7) = ab^7 = 38$

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

We now have two equations:

1. $ab^{0.5} = 16$
2. $ab^7 = 38$

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

$\frac{ab^7}{ab^{0.5}} = \frac{38}{16}$

Simplifying:

$b^{7 - 0.5} = \frac{38}{16} = 2.375$

This gives:

$b^{6.5} = 2.375$

To solve for $b$, take the 6.5th root of both sides:

$b = \left( 2.375 \right)^{\frac{1}{6.5}} \approx 1.138$

Step 3: Solve for $a$

Now that we have $b \approx 1.138$, substitute it back into the first equation:

$ab^{0.5} = 16$

$a \times 1.138^{0.5} = 16$

Simplifying $1.138^{0.5} \approx 1.067$, we get:

$a \times 1.067 = 16$

Solving for $a$:

$a \approx \frac{16}{1.067} \approx 15.0$

Step 4: Find $f(1)$

Now that we have $a \approx 15.0$ and $b \approx 1.138$, substitute into the function $f(x) = ab^x$ to find $f(1)$:

$f(1) = 15.0 \times 1.138^1 = 15.0 \times 1.138 \approx 17.07$

Final Answer:

The value of $f(1)$ to the nearest hundredth is:

$f(1) \approx 17.07$

Would you like more details or have any questions?

Here are 5 related questions:

1. What would happen if $f(x)$ were a linear function instead of exponential?
2. How can the rate of growth of an exponential function be interpreted from $b$?
3. What if we knew $f(x)$ at three points? How would that affect the process?
4. Can you find the derivative of $f(x) = ab^x$ with respect to $x$?
5. How does the base $b$ relate to exponential growth or decay?

Tip: When solving for unknown parameters in exponential functions, using logarithms can help simplify the calculations involving powers.