What is the average rate of change of the function on the interval from x = 0 to x = 5 ; f(x)= 1\2 (3)^x
Answer:
Average rate of change of the function on the interval from x = 0 to x = 5 is 24.2
Solution:
Given function is:
[tex]f(x) = \frac{1}{2}(3^x)[/tex]
We have to find the average rate of change of function from x = 0 to x = 5
The formula for average rate of change can be expressed as follows:
[tex]{A\left( x \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}}[/tex]
So for rate of change of function from x = 0 to x = 5 is:
[tex]{A\left( x \right) = \frac{{f\left( 5 \right) - f\left( 0 \right)}}{{5 - 0}}}[/tex]
Let us find f(0) and f(5)
To find f(0), substitute x = 0 in f(x)
[tex]f(0) = \frac{1}{2}(3^0) = \frac{1}{2}[/tex]
To find f(5), substitute x = 5 in f(x)
[tex]f(5) = \frac{1}{2}(3^5) = \frac{1}{2}(243) = \frac{243}{2}[/tex]
Therefore,
[tex]A(x)=\frac{\frac{243}{2}-\frac{1}{2}}{5-0}=\frac{242}{\frac{2}{5}}=\frac{242}{2} \times \frac{1}{5}=24.2[/tex]
Therefore average rate of change of the function on the interval from x = 0 to x = 5 is 24.2
