AP Calculus BC 5.3 Determining Intervals on Which a Function Is Increasing or Decreasing - FRQs - Exam Style Questions

(a)
On \([0,4]\), \(f’\) is a semicircle of radius \(2\) below the \(x\)-axis, so the signed area is \(-2\pi\).
\[ f(0)=f(4)-\int_{0}^{4} f'(x)\,dx=3-(-2\pi)=3+2\pi. \] From \(x=4\) to \(x=5\) the linear piece contributes area \(\tfrac12\). Hence \[ f(5)=f(4)+\int_{4}^{5} f'(x)\,dx=3+\tfrac12=\tfrac72. \]
(b)
Points of inflection of \(f\) occur where \(f”\) changes sign. On the graph of \(f’\), this is where the slope of \(f’\) changes from decreasing to increasing or vice versa. This happens at \(x=2\) (bottom of the semicircle) and \(x=6\) (peak of the linear rise).
\(\Rightarrow\) Inflection points at \(x=2\) and \(x=6\).
(c)
\[ g'(x)=f'(x)-1. \] \(g\) is decreasing where \(g'(x)\le 0\), i.e., where \(f'(x)\le 1\). From the graph this holds for \(0\le x\le 5\).
\(\Rightarrow\) \(g\) decreases on \([0,5]\).
(d)
Critical numbers for \(g\) satisfy \(g'(x)=0\Rightarrow f'(x)=1\). From the graph, this occurs at \(x=5\). Check endpoints and this critical point: \[ g(0)=f(0)-0=3+2\pi,\quad g(5)=f(5)-5=\tfrac72-5=-\tfrac32. \] From \(5\) to \(7\), the “up” and “down” triangle areas cancel, so \(f(7)=f(5)=\tfrac72\) and thus \(g(7)=\tfrac72-7=-\tfrac12\).
Since \(g'(x)<0\) on \((0,5)\) and \(g'(x)>0\) on \((5,7)\), the absolute minimum on \([0,7]\) is \[ \boxed{-\tfrac32}\ \text{at } x=5. \]