$<3 / 8>$ होiline for $y=f(x)=4 x^{\wedge} 3, x=5$, and $\Delta x=0.02$ find $\Delta y$ for the given $x$ and $\Delta x$ values

Step 1 ：The problem is asking for the change in y, denoted as \(\Delta y\), for the function \(y=f(x)=4x^3\) at \(x=5\) with a small change in x, denoted as \(\Delta x\), of 0.02.

Step 2 ：To find \(\Delta y\), we can use the formula for the derivative of a function, which gives us the rate of change of the function at a given point. The derivative of \(f(x)=4x^3\) is \(f'(x)=12x^2\).

Step 3 ：We can then use this derivative to find the approximate change in y for a small change in x, using the formula \(\Delta y \approx f'(x) * \Delta x\).

Step 4 ：So, we need to calculate \(f'(5) * 0.02\).

Step 5 ：The calculation gives us the value of \(\Delta y\) as 6.0. This is the approximate change in y for a small change in x of 0.02 at the point \(x=5\) on the function \(y=f(x)=4x^3\).

Step 6 ：Final Answer: \(\boxed{6.0}\)