Answer :
We start with the exponential function given by
[tex]$$
f(x) = a b^x.
$$[/tex]
Step 1. Find the constant [tex]\( a \)[/tex]:
Since [tex]\( f(0) = a \cdot b^0 = a \)[/tex], we are given
[tex]$$
f(0) = 82,
$$[/tex]
so
[tex]$$
a = 82.
$$[/tex]
Step 2. Find the constant [tex]\( b \)[/tex]:
We are also given
[tex]$$
f(-3.5) = 6.
$$[/tex]
Substitute [tex]\( x = -3.5 \)[/tex] into the function:
[tex]$$
f(-3.5) = 82 \cdot b^{-3.5} = 6.
$$[/tex]
Isolate [tex]\( b^{-3.5} \)[/tex] by dividing both sides by 82:
[tex]$$
b^{-3.5} = \frac{6}{82}.
$$[/tex]
It is more convenient to work with a positive exponent. Recall that
[tex]$$
b^{-3.5} = \frac{1}{b^{3.5}},
$$[/tex]
so we take the reciprocal:
[tex]$$
b^{3.5} = \frac{82}{6}.
$$[/tex]
Now, solve for [tex]\( b \)[/tex] by taking the [tex]\( \frac{1}{3.5} \)[/tex] power on both sides:
[tex]$$
b = \left(\frac{82}{6}\right)^{\frac{1}{3.5}}.
$$[/tex]
Calculating the approximate value, we obtain
[tex]$$
b \approx 2.1109358187.
$$[/tex]
Step 3. Calculate [tex]\( f(-0.5) \)[/tex]:
Now, we use the function with [tex]\( x = -0.5 \)[/tex]:
[tex]$$
f(-0.5) = 82 \cdot b^{-0.5}.
$$[/tex]
Substitute the value of [tex]\( b \)[/tex]:
[tex]$$
f(-0.5) \approx 82 \cdot \left(2.1109358187\right)^{-0.5}.
$$[/tex]
Evaluating this expression, we find
[tex]$$
f(-0.5) \approx 56.4386137210.
$$[/tex]
Step 4. Round to the nearest hundredth:
Rounding the result to the nearest hundredth gives
[tex]$$
f(-0.5) \approx 56.44.
$$[/tex]
Thus, the value of [tex]\( f(-0.5) \)[/tex] is approximately
[tex]$$
\boxed{56.44}.
$$[/tex]
[tex]$$
f(x) = a b^x.
$$[/tex]
Step 1. Find the constant [tex]\( a \)[/tex]:
Since [tex]\( f(0) = a \cdot b^0 = a \)[/tex], we are given
[tex]$$
f(0) = 82,
$$[/tex]
so
[tex]$$
a = 82.
$$[/tex]
Step 2. Find the constant [tex]\( b \)[/tex]:
We are also given
[tex]$$
f(-3.5) = 6.
$$[/tex]
Substitute [tex]\( x = -3.5 \)[/tex] into the function:
[tex]$$
f(-3.5) = 82 \cdot b^{-3.5} = 6.
$$[/tex]
Isolate [tex]\( b^{-3.5} \)[/tex] by dividing both sides by 82:
[tex]$$
b^{-3.5} = \frac{6}{82}.
$$[/tex]
It is more convenient to work with a positive exponent. Recall that
[tex]$$
b^{-3.5} = \frac{1}{b^{3.5}},
$$[/tex]
so we take the reciprocal:
[tex]$$
b^{3.5} = \frac{82}{6}.
$$[/tex]
Now, solve for [tex]\( b \)[/tex] by taking the [tex]\( \frac{1}{3.5} \)[/tex] power on both sides:
[tex]$$
b = \left(\frac{82}{6}\right)^{\frac{1}{3.5}}.
$$[/tex]
Calculating the approximate value, we obtain
[tex]$$
b \approx 2.1109358187.
$$[/tex]
Step 3. Calculate [tex]\( f(-0.5) \)[/tex]:
Now, we use the function with [tex]\( x = -0.5 \)[/tex]:
[tex]$$
f(-0.5) = 82 \cdot b^{-0.5}.
$$[/tex]
Substitute the value of [tex]\( b \)[/tex]:
[tex]$$
f(-0.5) \approx 82 \cdot \left(2.1109358187\right)^{-0.5}.
$$[/tex]
Evaluating this expression, we find
[tex]$$
f(-0.5) \approx 56.4386137210.
$$[/tex]
Step 4. Round to the nearest hundredth:
Rounding the result to the nearest hundredth gives
[tex]$$
f(-0.5) \approx 56.44.
$$[/tex]
Thus, the value of [tex]\( f(-0.5) \)[/tex] is approximately
[tex]$$
\boxed{56.44}.
$$[/tex]
