Question

webwork / mt1303abd fa25 / hw 15 exponential functions 6.1 / 10
previous problem problem list next problem
hw 15 exponential functions 6.1: problem 10
(1 point)
for the function $f(x) = 2.4(3)^{-x + 1} + 2.5$, find $f(3)$. round answer to four decimal places, if necessary.
$f(3) = \square$ help (numbers)
preview my answers submit answers show me another
you have attempted this problem 0 times. you have 5 attempts left before new version will be requested.
you have unlimited attempts remaining.

Explanation:

Step1: Substitute \(x = 3\) into the function

Given \(f(x)=2.4(3)^{-x + 1}+2.5\), when \(x = 3\), we have \(f(3)=2.4(3)^{-3 + 1}+2.5\).
First, calculate the exponent: \(-3+1=-2\). So \(f(3)=2.4(3)^{-2}+2.5\).

Step2: Simplify the exponential term

Using the rule \(a^{-n}=\frac{1}{a^{n}}\), then \(3^{-2}=\frac{1}{3^{2}}=\frac{1}{9}\).
So \(f(3)=2.4\times\frac{1}{9}+2.5\).

Step3: Calculate the product and sum

\(2.4\times\frac{1}{9}=\frac{2.4}{9}\approx0.2667\).
Then \(f(3)=0.2667 + 2.5\).
\(f(3)=2.7667\).

Answer:

\(2.7667\)