Given the graph of a continuous function f (two semicircles and a line segment), students analyze an accumulation function g: find a derivative value, locate the points of inflection of g, evaluate g at points using signed geometric area, and find where g attains its absolute minimum on a closed interval.
Given the graph of a continuous function f (two semicircles and a line segment), students analyze an accumulation function g: find a derivative value, locate the points of inflection of g, evaluate g at points using signed geometric area, and find where g attains its absolute minimum on a closed interval.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
Since $g(x)=\int_{6}^{x} f(t)\,dt$, the Fundamental Theorem of Calculus (Part 1) gives $g'(x)=f(x)$. Therefore $$g'(8)=f(8).$$ From the figure, on the interval $[6,12]$ the graph of $f$ is the line segment from $(6,0)$ to $(12,3)$, which has slope $\tfrac{3-0}{12-6}=\tfrac12$, so $f(8)=\tfrac12(8-6)=1$. Thus $$g'(8)=f(8)=1.$$
1 point for recognizing g'(x)=f(x) (the FTC application); 1 point for the value g'(8)=f(8)=1 read from the graph. A response that jumps straight to g'(8)=1 via an implied FTC can still earn the answer point; computing a difference of f-values (e.g. f(8)-f(6)) earns the value point but not the FTC-reasoning point.
The graph of $g$ has a point of inflection where $g''=f'$ changes sign, i.e. where $f$ changes from increasing to decreasing or vice versa. Reading the figure: - At $x=-3$ (bottom of the lower semicircle): $f$ changes from decreasing to increasing $\Rightarrow$ inflection. - At $x=3$ (top of the upper semicircle): $f$ changes from increasing to decreasing $\Rightarrow$ inflection. - At $x=6$ (where the upper semicircle meets the line segment): $f$ changes from decreasing to increasing $\Rightarrow$ inflection. So $g$ has points of inflection at $x=-3,\ 3,\ \text{and}\ 6$.
1 point (answer) requires exactly x=-3, x=3, and x=6 with no extra/incorrect values in the open interval; 1 point (reason) requires tying the inflection to the graph of f changing from increasing to decreasing (or vice versa) -- equivalently, the slope of f changing sign, or f attaining relative extrema there. Citing g'' = f' changing sign without referencing the given graph of f earns the answer point but not the reason point.
Use signed geometric areas under $f$, with the lower integration limit $6$. **$g(12)$:** On $[6,12]$, $f$ is the line segment from $(6,0)$ to $(12,3)$, forming a triangle above the axis with base $6$ and height $3$: $$g(12)=\int_{6}^{12} f(t)\,dt=\tfrac12(6)(3)=9.$$ **$g(0)$:** Reverse the limits. On $[0,6]$, $f$ is the upper semicircle of radius $3$, area $=\tfrac12\pi(3)^2=\tfrac{9\pi}{2}$ (above the axis, positive). So $$g(0)=\int_{6}^{0} f(t)\,dt=-\int_{0}^{6} f(t)\,dt=-\frac{9\pi}{2}.$$
1 point for g(12)=9 (triangle area, labeled); 1 point for g(0) = -9*pi/2 (semicircle area with the sign from reversing the limits, labeled). Unlabeled values earn neither point. Each value is awarded with or without supporting work as long as it is correctly labeled to g(12) or g(0).
On the closed interval $-6\le x\le 12$, $g$ attains its absolute minimum either where $g'(x)=f(x)=0$ or at an endpoint (candidates test). From the figure, $f(x)=0$ at $x=0$ and $x=6$, so the candidates are $x=-6,\,0,\,6,\,12$. Evaluate $g$ at each candidate (using signed areas, limits from $6$): | $x$ | $g(x)$ | |---|---| | $-6$ | $0$ | | $0$ | $-\dfrac{9\pi}{2}$ | | $6$ | $0$ | | $12$ | $9$ | Note $g(-6)=\int_{6}^{-6} f\,dt = 0$ because the lower semicircle on $[-6,0]$ (area $-\tfrac{9\pi}{2}$) and the upper semicircle on $[0,6]$ (area $+\tfrac{9\pi}{2}$) cancel. The smallest of these values is $-\dfrac{9\pi}{2}\approx -14.137$ at $x=0$. Therefore $g$ attains its absolute minimum at $x=0$.
1 point (P7) for considering g'(x)=0 / f(x)=0 (the critical points), or equivalently analyzing the sign of g'=f; 1 point (P8) for a complete global justification -- a candidates test evaluating g(-6), g(0), g(6), g(12) (and no other x-values), or the sign argument f(x)<=0 on [-6,0] and f(x)>=0 on [0,12]; 1 point (P9) for the answer x=0. P9 can be earned with values imported from part (c). A purely local first/second-derivative-test argument does not earn P8 but is still eligible for P9.
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