AP Calc FRQ Set #2 Study Guide

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Last updated 2:31 AM on 4/16/26
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6 Terms

1
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<p>2000 AB 4</p>

2000 AB 4

c) A(t) = 30 +∫t0 8 - sqrt(x+1) dx

A(t) = 30 + 8t - ∫t0 sqrt(x+1) dx

d) A’(t) = 8 - sqrt (x-1); A(t) = 0 when t = 63

A’(t) is positive for (0, 63) but negative for (63, 120); therefore, there is an absolute maximum at t=63.

<p>c) A(t) = 30 +<strong>∫<sup>t</sup><sub>0</sub> </strong>8 - sqrt(x+1) dx</p><p>A(t) = 30 + 8t - <strong>∫<sup>t</sup><sub>0</sub> </strong>sqrt(x+1) dx</p><p></p><p>d) A’(t) = 8 - sqrt (x-1); A(t) = 0 when t = 63</p><p>A’(t) is positive for (0, 63) but negative for (63, 120); therefore, there is an absolute maximum at t=63.</p>
2
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<p>2007 AB 3</p>

2007 AB 3

d) establish that since g(1) = 2), g-1 (2) = 1

g-1’(x) = 1/g’(g-1(x))

<p>d) establish that since g(1) = 2), g<sup>-1</sup> (2) = 1</p><p>g<sup>-1’</sup>(x) = 1/g’(g<sup>-1</sup>(x))</p>
3
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<p>2017 ABBC 3</p>

2017 ABBC 3

b) increasing on [-6, -2] u [2,5] since f’(x) > 0 for [-6, -2) u (2, 5)

d) Find limit on left and right, but then use slope formula with generic point as first and f’(3) as second to prove at end.

<p>b) increasing on [-6, -2] u [2,5] since f’(x) &gt; 0 for [-6, -2) u (2, 5)</p><p>d) Find limit on left and right, but then use slope formula with generic point as first and f’(3) as second to prove at end. </p>
4
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<p>2019 AB 2 CALC</p>

2019 AB 2 CALC

a) interval [0.3, 3.8] is continous.

c) technically taking intergral of absolute value of vqt

<p>a) interval [0.3, 3.8] is continous.</p><p>c) technically taking intergral of absolute value of v<sub>q</sub>t</p>
5
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<p>2022 AB 2 CALC</p>

2022 AB 2 CALC

a) Make sure you state equations equal each other, and then B

d) d/dt (A(x)) = DA/dx (dx/dt)

D/dt (A(x))| x=-0.5 = A’(-.0.5)(7) = -9.272

<p>a) Make sure you state equations equal each other, and then B</p><p>d) d/dt (A(x)) = DA/dx (dx/dt)</p><p>D/dt (A(x))| <sub>x=-0.5<sup> </sup></sub>= A’(-.0.5)(7) = -9.272</p>
6
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<p>2022 AB 5</p>

2022 AB 5

d) Coefficient of cos is -1/pi

<p>d) Coefficient of cos is -1/pi</p>