Given a function f(x) = x(9 - x).
We need to find the value of f(2) and f(2.1) and use them to approximate the value of f'(2).
The value of f(2) is calculated below:
[tex]\begin{gathered} f(2)=2(9-2) \\ =2(7) \\ =14 \end{gathered}[/tex]The value of the f(2.1) is calculated as follows:
[tex]\begin{gathered} f(2.1)=2.1(9-2.1) \\ =2.1(6.9) \\ =14.49 \end{gathered}[/tex]Now, by the definition of f'(x), we know that
[tex]f^{\prime}(x)=\frac{f(x+\Delta x)-f(x)}{(x+\Delta x)-x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}[/tex]For the given condition, x = 2, and delta x = 0.1. So, the value of f'(2) is
[tex]\begin{gathered} f^{\prime}(2)=\frac{f(2+0.1)-f(2)}{0.1} \\ =\frac{f(2.1)-f(2)}{0.1} \\ =\frac{14.49-14}{0.1} \\ =\frac{0.49}{0.1} \\ =4.9 \end{gathered}[/tex]Thus, the approximate value of f'(2) is 4.9.
