The graph of f' (built from line segments and a semicircle) is given for a function f with a known value. Students classify an extremum of f and identify concavity from the behavior of f', evaluate an indeterminate-form limit, and determine the absolute minimum of f on a closed interval using accumulated signed area.
The graph of f' (built from line segments and a semicircle) is given for a function f with a known value. Students classify an extremum of f and identify concavity from the behavior of f', evaluate an indeterminate-form limit, and determine the absolute minimum of f on a closed interval using accumulated signed area.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
A relative extremum of $f$ occurs only where $f'$ **changes sign**. From the graph of $f'$, we have $f'(x)>0$ on $(2,6)$ and also $f'(x)>0$ on $(6,8)$. Even though $f'(6)=0$ (the semicircle meets the axis there), $f'$ stays positive on both sides of $x=6$ — it does **not** change sign. Therefore $f$ has **neither a relative minimum nor a relative maximum at $x=6$.**
1 point. Earned by stating that f' does not change sign at x=6, so neither. A response need not list sign intervals, but any listed must be correct (endpoints optional). Merely stating 'f'>0 before and after x=6' without the 'neither' conclusion does not earn the point.
The graph of $f$ is concave down exactly where $f'$ is **decreasing** (since $f''=(f')'<0$ there). Reading the graph of $f'$, it is decreasing on the intervals where its segments/semicircle go downward: $$f' \text{ is decreasing on } (-2,0) \text{ and on } (4,6).$$ Therefore the graph of $f$ is **concave down on $(-2,0)$ and $(4,6)$**, because $f'$ is decreasing on these open intervals.
2 points. P1 (intervals): only (-2,0) and (4,6) (endpoints optional). P2 (reason): requires P1, plus correctly discussing that f' is decreasing (or the slopes of f' are decreasing) on those intervals. Special case: exactly one correct interval with correct reason earns 1 of 2.
Evaluate $\displaystyle\lim_{x\to 2}\frac{6f(x)-3x}{x^2-5x+6}$. Because $f$ is differentiable at $x=2$ it is continuous there, so $\lim_{x\to2}f(x)=f(2)=1$. Check the form: $$\lim_{x\to2}\big(6f(x)-3x\big)=6(1)-3(2)=0,\qquad \lim_{x\to2}\big(x^2-5x+6\big)=4-10+6=0.$$ This is the indeterminate form $\frac{0}{0}$, so **L'Hospital's Rule** applies. Differentiating numerator and denominator: $$\lim_{x\to2}\frac{6f(x)-3x}{x^2-5x+6}=\lim_{x\to2}\frac{6f'(x)-3}{2x-5}=\frac{6f'(2)-3}{2(2)-5}=\frac{6(0)-3}{-1}=\boxed{3}.$$ Here $f'(2)=0$ is read from the graph of $f'$.
3 points. P1 (limits of numerator and denominator): present two separate limits each equal to 0; writing the quotient as 0/0 directly does NOT earn P1. P2 (uses L'Hospital): present at least one correct derivative in the ratio of derivatives. P3 (answer): the value 3 with supporting work. Uses f'(2)=0 and f(2)=1.
On $[-2,8]$, the candidates for an absolute minimum of $f$ are the critical points (where $f'=0$) and the endpoints. From the graph $f'(x)=0$ at $x=-1,\,2,\,6$, and $f$ is continuous on the closed interval, so the candidates are $x=-2,-1,2,6,8$. Using $f(2)=1$ and the **signed area** under $f'$ (Fundamental Theorem of Calculus, $f(b)-f(a)=\int_a^b f'\,dt$), evaluate $f$ at each candidate: | $x$ | $f(x)$ | |---|---| | $-2$ | $3$ | | $-1$ | $4$ | | $2$ | $1$ | | $6$ | $7-\pi\approx 3.858$ | | $8$ | $11-2\pi\approx 4.717$ | The smallest of these values is $f(2)=1$. Therefore the **absolute minimum value of $f$ on $[-2,8]$ is $\boxed{1}$** (attained at $x=2$).
3 points. P1 (considers f'=0): state f'=0 or equivalent; merely listing the zeros of f' is not sufficient. P2 (justification): a candidate-test comparison; any error evaluating f at a critical point or endpoint forfeits P2, and both endpoints must be considered. P3 (answer): the minimum VALUE is 1 (not merely 'at x=2'). x=-1 may be dropped as a local max and x=6 dropped via part (a); f(8)>f(2) may be argued from f'≥0 for x>2.
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