A table gives values of two twice-differentiable functions and their first derivatives at selected inputs. Students compute the derivative of a composite function, judge the concavity of a product-defined function via its second derivative, evaluate an accumulation-function value, and determine whether that function is increasing or decreasing at a point.
A table gives values of two twice-differentiable functions and their first derivatives at selected inputs. Students compute the derivative of a composite function, judge the concavity of a product-defined function via its second derivative, evaluate an accumulation-function value, and determine whether that function is increasing or decreasing at a point.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
With $h(x)=f(g(x))$, the **chain rule** gives $h'(x)=f'\big(g(x)\big)\cdot g'(x)$. Evaluate at $x=7$ using the table $g(7)=0$, $g'(7)=8$, and $f'(0)=\tfrac{3}{2}$: $$h'(7)=f'\big(g(7)\big)\cdot g'(7)=f'(0)\cdot 8=\frac{3}{2}\cdot 8=\boxed{12}.$$
2 points. P1 (chain rule): present h'(x)=f'(g(x))·g'(x) or h'(7)=f'(g(7))·g'(7). P2 (answer): 12, earned only if P1 is earned, OR (if P1 not earned) for f'(0)·8=12 or (3/2)·8. A bare 12 with no supporting work earns neither point.
Here $k'(x)=\big(f(x)\big)^2\cdot g(x)$. Differentiate again with the **product and chain rules**: $$k''(x)=2f(x)\,f'(x)\,g(x)+\big(f(x)\big)^2 g'(x).$$ Evaluate at $x=4$ with $f(4)=4$, $f'(4)=3$, $g(4)=-3$, $g'(4)=2$: $$k''(4)=2(4)(3)(-3)+(4)^2(2)=-72+32=\boxed{-40}.$$ Since $k''(4)=-40<0$ (and $k''$ is continuous), the graph of $k$ is **concave down** at $x=4$.
3 points. P1 (product/chain rule): present k''(x)=2f f' g + f^2 g' (or evaluated at 4); also earned by the SG's listed single-error forms (which then cannot earn P2). P2 (value): k''(4)=-40 with supporting work. P3 (answer with reason): concave down because k''(4)<0, consistent with any declared nonzero k''(4).
With $m(x)=5x^3+\displaystyle\int_0^x f'(t)\,dt$, evaluate at $x=2$. The integral is handled by the **Fundamental Theorem of Calculus**: $\int_0^2 f'(t)\,dt=f(2)-f(0)$. $$m(2)=5(2)^3+\int_0^2 f'(t)\,dt=5(8)+\big(f(2)-f(0)\big)=40+(7-10)=40-3=\boxed{37}.$$
1 point. Earned only for the answer 37 with supporting work equivalent to 5·8+(f(2)-f(0)), 40+(f(2)-f(0)), 5·8+(7-10), or 40+(7-10). A bare 37 with no supporting work does not earn the point.
Differentiate $m(x)=5x^3+\int_0^x f'(t)\,dt$. By the Fundamental Theorem of Calculus, $\frac{d}{dx}\int_0^x f'(t)\,dt=f'(x)$, so $$m'(x)=15x^2+f'(x).$$ Evaluate at $x=2$ using $f'(2)=-8$: $$m'(2)=15(2)^2+f'(2)=60+(-8)=52.$$ Since $m'(2)=52>0$, the function $m$ is **increasing at $x=2$.**
3 points. P1 (considers m'): present m'(x), m'(2), or m' (may appear in the justification). P2 (m'(2) with supporting work): 15·2^2+f'(2), 60+f'(2), or 60-8; an unsupported m'(2)=52 does not earn P2. P3 (answer with justification): increasing because m'(2)>0, consistent with any declared value of m'(2).
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