We can use the first derivative test to determine the local minimum and maximum of A(x) on the interval (0,6).
Taking the derivative of A(x) with respect to x, we get:
A'(x) = f(x)
Note that A(x) is the integral of f(x), so A'(x) is just f(x) by the fundamental theorem of calculus.
To find the critical points of A(x) on (0,6), we set A'(x) = 0 and solve for x:
f(x) = 0
From the given figure, we can see that f(x) is positive on the interval (0,2) and negative on the interval (2,6). Therefore, there must be a local maximum of A(x) at some point in (2,6) where f(x) = 0.
To find the local minimum of A(x) on (0,6), we need to look for a point where A'(x) changes sign from positive to negative. From the given figure, we can see that f(x) is increasing on the interval (0,1) and decreasing on the interval (1,6). Therefore, A'(x) = f(x) is also increasing on (0,1) and decreasing on (1,6). This means that A(x) has a local minimum at the point x where A'(x) = f(x) changes sign from positive to negative. From the figure, we can see that this occurs at x = 2.
Therefore, the local minimum of A(x) on (0,6) is at x = 2, and the local maximum of A(x) on (0,6) is at some point in (2,6) where f(x) = 0.