Answer :
We start with the exponential function
[tex]$$
f(x) = a \, b^x.
$$[/tex]
Since [tex]\( f(0) = 84 \)[/tex], we have
[tex]$$
f(0) = a \, b^0 = a = 84.
$$[/tex]
Next, using the point [tex]\( f(-5) = 12 \)[/tex], we substitute [tex]\( x = -5 \)[/tex] into the function:
[tex]$$
f(-5) = 84 \, b^{-5} = 12.
$$[/tex]
Dividing both sides of the equation by 84 gives
[tex]$$
b^{-5} = \frac{12}{84} = \frac{1}{7}.
$$[/tex]
Taking the reciprocal of both sides, we obtain
[tex]$$
b^5 = 7.
$$[/tex]
Thus, the base [tex]\( b \)[/tex] is
[tex]$$
b = 7^{\frac{1}{5}}.
$$[/tex]
Now, to find [tex]\( f(-3.5) \)[/tex], substitute [tex]\( x = -3.5 \)[/tex] into the function:
[tex]$$
f(-3.5) = 84 \, b^{-3.5}.
$$[/tex]
Replacing [tex]\( b \)[/tex] with [tex]\( 7^{\frac{1}{5}} \)[/tex]:
[tex]$$
f(-3.5) = 84 \left(7^{\frac{1}{5}}\right)^{-3.5} = 84 \, 7^{-\frac{3.5}{5}}.
$$[/tex]
Simplify the exponent:
[tex]$$
-\frac{3.5}{5} = -0.7,
$$[/tex]
so
[tex]$$
f(-3.5) = 84 \, 7^{-0.7}.
$$[/tex]
Evaluating [tex]\( f(-3.5) \)[/tex] numerically, we obtain approximately
[tex]$$
f(-3.5) \approx 21.51.
$$[/tex]
Thus, the value of [tex]\( f(-3.5) \)[/tex], rounded to the nearest hundredth, is
[tex]$$
\boxed{21.51}.
$$[/tex]
[tex]$$
f(x) = a \, b^x.
$$[/tex]
Since [tex]\( f(0) = 84 \)[/tex], we have
[tex]$$
f(0) = a \, b^0 = a = 84.
$$[/tex]
Next, using the point [tex]\( f(-5) = 12 \)[/tex], we substitute [tex]\( x = -5 \)[/tex] into the function:
[tex]$$
f(-5) = 84 \, b^{-5} = 12.
$$[/tex]
Dividing both sides of the equation by 84 gives
[tex]$$
b^{-5} = \frac{12}{84} = \frac{1}{7}.
$$[/tex]
Taking the reciprocal of both sides, we obtain
[tex]$$
b^5 = 7.
$$[/tex]
Thus, the base [tex]\( b \)[/tex] is
[tex]$$
b = 7^{\frac{1}{5}}.
$$[/tex]
Now, to find [tex]\( f(-3.5) \)[/tex], substitute [tex]\( x = -3.5 \)[/tex] into the function:
[tex]$$
f(-3.5) = 84 \, b^{-3.5}.
$$[/tex]
Replacing [tex]\( b \)[/tex] with [tex]\( 7^{\frac{1}{5}} \)[/tex]:
[tex]$$
f(-3.5) = 84 \left(7^{\frac{1}{5}}\right)^{-3.5} = 84 \, 7^{-\frac{3.5}{5}}.
$$[/tex]
Simplify the exponent:
[tex]$$
-\frac{3.5}{5} = -0.7,
$$[/tex]
so
[tex]$$
f(-3.5) = 84 \, 7^{-0.7}.
$$[/tex]
Evaluating [tex]\( f(-3.5) \)[/tex] numerically, we obtain approximately
[tex]$$
f(-3.5) \approx 21.51.
$$[/tex]
Thus, the value of [tex]\( f(-3.5) \)[/tex], rounded to the nearest hundredth, is
[tex]$$
\boxed{21.51}.
$$[/tex]
