Your function is piecewise continuous!
The first piece is just a straight line, y=x1, for 0≤x<3.
If the second piece really is a quarter circle, you will need its equation. The equation for a circle centered at (3,0) of radius 2 is (x3)² + y² = 4, so the quarter circle as a function is
y = sqrt(4(x3)²) = sqrt(x²+6x5), for 3≤x≤5.
So the piecewise function g(t) is
g(t) = { (t1) 0≤t<3
{ sqrt(t²+6t5) 3≤t≤5
Now f(x)=∫_{0}^{x} g(t) dt, which will also be piecewise continuous. It's the area between the graph of g(t) and the taxis, counting areas below the taxis as negative. To find the absolute extrema of f(x), you do not need to carry out
the integral, assuming you know the area of a circle and of a triangle. For which x is the area smallest? For which x is the area largest? What are the min and max values of these areas?
Nov 25

Andre W.
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