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Answer :
We start with the exponential function
[tex]$$
f(x) = a b^x,
$$[/tex]
and we are given the two conditions
[tex]$$
f(-1) = 2 \quad \text{and} \quad f(2) = 84.
$$[/tex]
Step 1. Express one parameter in terms of the other
From the condition at [tex]$x=-1$[/tex], we have
[tex]$$
a \cdot b^{-1} = 2.
$$[/tex]
This can be rearranged to express [tex]$a$[/tex] in terms of [tex]$b$[/tex]:
[tex]$$
a = 2b.
$$[/tex]
Step 2. Substitute into the second condition to find [tex]$b$[/tex]
Now use the condition at [tex]$x=2$[/tex]:
[tex]$$
f(2) = a \cdot b^2 = 84.
$$[/tex]
Substitute [tex]$a = 2b$[/tex] into this equation:
[tex]$$
(2b) \cdot b^2 = 84 \quad \Longrightarrow \quad 2b^3 = 84.
$$[/tex]
Divide both sides by 2:
[tex]$$
b^3 = 42.
$$[/tex]
Taking the cube root of both sides gives
[tex]$$
b = 42^{\frac{1}{3}}.
$$[/tex]
Step 3. Find [tex]$a$[/tex]
Recall that
[tex]$$
a = 2b,
$$[/tex]
so substituting the value we found for [tex]$b$[/tex]:
[tex]$$
a = 2 \times 42^{\frac{1}{3}}.
$$[/tex]
Step 4. Calculate [tex]$f(2.5)$[/tex]
We now need to evaluate
[tex]$$
f(2.5) = a \cdot b^{2.5}.
$$[/tex]
Substitute [tex]$a = 2b$[/tex] into the equation:
[tex]$$
f(2.5) = (2b) \cdot b^{2.5} = 2 \cdot b^{3.5}.
$$[/tex]
Using the numerical approximation [tex]$b \approx 3.47602664488645$[/tex], we get
[tex]$$
f(2.5) \approx 2 \cdot (3.47602664488645)^{3.5} \approx 156.61048498206875.
$$[/tex]
Rounding this to the nearest hundredth, we find
[tex]$$
f(2.5) \approx 156.61.
$$[/tex]
Thus, the value of [tex]$f(2.5)$[/tex], rounded to the nearest hundredth, is [tex]$\boxed{156.61}$[/tex].
[tex]$$
f(x) = a b^x,
$$[/tex]
and we are given the two conditions
[tex]$$
f(-1) = 2 \quad \text{and} \quad f(2) = 84.
$$[/tex]
Step 1. Express one parameter in terms of the other
From the condition at [tex]$x=-1$[/tex], we have
[tex]$$
a \cdot b^{-1} = 2.
$$[/tex]
This can be rearranged to express [tex]$a$[/tex] in terms of [tex]$b$[/tex]:
[tex]$$
a = 2b.
$$[/tex]
Step 2. Substitute into the second condition to find [tex]$b$[/tex]
Now use the condition at [tex]$x=2$[/tex]:
[tex]$$
f(2) = a \cdot b^2 = 84.
$$[/tex]
Substitute [tex]$a = 2b$[/tex] into this equation:
[tex]$$
(2b) \cdot b^2 = 84 \quad \Longrightarrow \quad 2b^3 = 84.
$$[/tex]
Divide both sides by 2:
[tex]$$
b^3 = 42.
$$[/tex]
Taking the cube root of both sides gives
[tex]$$
b = 42^{\frac{1}{3}}.
$$[/tex]
Step 3. Find [tex]$a$[/tex]
Recall that
[tex]$$
a = 2b,
$$[/tex]
so substituting the value we found for [tex]$b$[/tex]:
[tex]$$
a = 2 \times 42^{\frac{1}{3}}.
$$[/tex]
Step 4. Calculate [tex]$f(2.5)$[/tex]
We now need to evaluate
[tex]$$
f(2.5) = a \cdot b^{2.5}.
$$[/tex]
Substitute [tex]$a = 2b$[/tex] into the equation:
[tex]$$
f(2.5) = (2b) \cdot b^{2.5} = 2 \cdot b^{3.5}.
$$[/tex]
Using the numerical approximation [tex]$b \approx 3.47602664488645$[/tex], we get
[tex]$$
f(2.5) \approx 2 \cdot (3.47602664488645)^{3.5} \approx 156.61048498206875.
$$[/tex]
Rounding this to the nearest hundredth, we find
[tex]$$
f(2.5) \approx 156.61.
$$[/tex]
Thus, the value of [tex]$f(2.5)$[/tex], rounded to the nearest hundredth, is [tex]$\boxed{156.61}$[/tex].
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