Difference between revisions of "Midterm information/Midterm 2015/Commentary"
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− | <li> It is possible to arrive at the correct answer (3/2) using an incorrect formula for the linear approximation. The incorrect formula that "works" is $f(2)\approx f'(2)+f(2)(2-1)$. This would get you 1 point simply for having the right idea for how to proceed even though you got the formula wrong. If you got a point for an answer of 3/2 by this incorrect method, DO NOT get the impression that you got the point for getting the numerical answer right. </li> | + | <li> It is possible to arrive at the correct answer (3/2) using an incorrect formula for the linear approximation. The incorrect formula that "works" is $f(2)\approx f'(2)+f(2)(2-1)$. This would get you 1 point simply for having the right idea for how to proceed even though you got the formula wrong. If you got a point for an answer of 3/2 by this incorrect method, DO NOT get the impression that you got the point for getting the numerical answer right. Another formula error that got a single point (despite not getting an answer of 3/2) was $f(2)\approx f(2)-f'(2)(2-1)$.</li> |
<li>Mentioning concave up is not sufficient to get a point. Concavity must be linked to the approximation being an underestimate to get the first point.</li> | <li>Mentioning concave up is not sufficient to get a point. Concavity must be linked to the approximation being an underestimate to get the first point.</li> | ||
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Revision as of 21:17, 25 October 2015
Here are some clarifications of the marking scheme for the midterm that are not included in the pdf file.
- Each box is worth one point. Putting $2x^3+1$ AND $x^2+2$ in the first two boxes would get a single point for recognizing which part of the function to use for each limit but failing to calculate the limit.
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- It is possible to arrive at the correct answer (3/2) using an incorrect formula for the linear approximation. The incorrect formula that "works" is $f(2)\approx f'(2)+f(2)(2-1)$. This would get you 1 point simply for having the right idea for how to proceed even though you got the formula wrong. If you got a point for an answer of 3/2 by this incorrect method, DO NOT get the impression that you got the point for getting the numerical answer right. Another formula error that got a single point (despite not getting an answer of 3/2) was $f(2)\approx f(2)-f'(2)(2-1)$.
- Mentioning concave up is not sufficient to get a point. Concavity must be linked to the approximation being an underestimate to get the first point.
- Providing a function $g(x)$ that has a root in the correct location ($25^{2/3}$) gets you a single point unless it is a polynomial (polynomials must have INTEGER powers). Stating $x_0=3$ should get you a point for any $g(x)$. If your $g(x)$ included $\sqrt{x}$ somewhere in it, then $x_0=4$ would also get you a point.
- You get two points for getting the horizontal and vertical coordinates of the min/max in the right location (one point for each coordinate). $x_{max}$ should be between -40 and -24. It should be lined up with the inflection point of $N(x)$. Also, $N'(x_{max})$ should be between 4 and 6. There was also a point for showing $N'(0)=0$. Another point could be removed if your graph came down to zero too early. The slope should not hit zero near or above x=-80 nor near or below x=80. We were lenient about the concavity of $N'(x)$ since it is a bit difficult to discern from $N(x)$ - even a piecewise linear graph could get a lot of points, provided all other details were correct.
- The graph should start above zero and should be smooth (no corners) and flat at t=0, t=1 and t=2. It should not consist of straight lines. The explanations should make reference to Lucas and his father, not the graph. Simply stating what is effectively a definition of a min, max or IP is not enough. The crucial facts that you should demonstrate awareness of are that (1) whenever Lucas' directions of travel (tangent line) is perpendicular to the line between him and his father, in an instantaneous sense, he is neither approaching nor moving away from his father and so the graph must have a zero slope there and (2) whenever Lucas' directions of travel (tangent line) is parallel to the line between him and his father, he is moving directly toward or away his father at a maximal rate meaning that the full extent of his constant speed is contributing to the change in distance to his father and so the graph has an inflection point there.