High School

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------------------------------------------------ If [tex]f(x)[/tex] is an exponential function of the form [tex]y = a b^x[/tex], where [tex]f(-5) = 12[/tex] and [tex]f(0) = 84[/tex], then find the value of [tex]f(-3.5)[/tex], to the nearest hundredth.

Answer :

We are given that the function has the form

[tex]$$
f(x) = a\,b^x,
$$[/tex]

with two conditions:

1. [tex]\( f(0) = 84 \)[/tex]
2. [tex]\( f(-5) = 12 \)[/tex]

Step 1. Find [tex]\( a \)[/tex]:

Since [tex]\( f(0) = a \, b^0 = a \)[/tex] and [tex]\( f(0) = 84 \)[/tex], we have

[tex]$$
a = 84.
$$[/tex]

Step 2. Find [tex]\( b \)[/tex]:

Using the condition [tex]\( f(-5) = 84 \, b^{-5} = 12 \)[/tex], we can solve for [tex]\( b \)[/tex]:

[tex]$$
84 \, b^{-5} = 12.
$$[/tex]

Divide both sides by 84:

[tex]$$
b^{-5} = \frac{12}{84} = \frac{1}{7}.
$$[/tex]

Recall that [tex]\( b^{-5} = \frac{1}{b^5} \)[/tex]. Hence, we have:

[tex]$$
\frac{1}{b^5} = \frac{1}{7}.
$$[/tex]

Taking the reciprocal of both sides gives:

[tex]$$
b^5 = 7.
$$[/tex]

Now, taking the fifth root of both sides:

[tex]$$
b = 7^{1/5}.
$$[/tex]

Step 3. Calculate [tex]\( f(-3.5) \)[/tex]:

Now, substitute [tex]\( a = 84 \)[/tex] and [tex]\( b = 7^{1/5} \)[/tex] into the function [tex]\( f(x) \)[/tex] for [tex]\( x = -3.5 \)[/tex]:

[tex]$$
f(-3.5) = 84 \, \left(7^{1/5}\right)^{-3.5}.
$$[/tex]

This expression simplifies the function evaluation. When computed, the value of [tex]\( f(-3.5) \)[/tex] (rounded to the nearest hundredth) is approximately

[tex]$$
21.51.
$$[/tex]

Thus, the final answer is

[tex]$$
\boxed{21.51}.
$$[/tex]