Calculus I Final Exam Solutions |

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**B/ Derivatives**

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**C/ Curve Sketching**

**D/ Max/Min**

10)

Now that we know

we know

Setting V'(r) = 0 will give us the critical points.

When r = 14/3 cm the cylinder will have a maximum volume.

The height will be 5 cm.

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**E/ Increments and Differentials**

**F/ Related Rates**

**G/ Mean Value Theorem**

13) Since **P** is the vertex of *f*, at (*2, 3*) and

since we move 36 units horizontally, **Q** is at
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**T** is the mean value point -- the point where the tangent is parallel to the chord.

So, since *b - a = 36* , and *f(b) - f(a)* = 3/4, then *36 f '(c)* must = 3/4.

Now, *
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When we solve for *c*, we get *c = 11* and *f(11) = 27/8*. Point **T** is at
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**H/ Antiderivatives**

14)

a) F(x) = 3x^{ 3 / 2} + 2x^{ 1 / 2} + C |
b) F(x) = 2/3 sin 3x + 3/2 cos 2x + C |

c) F(x) = 24/5 x^{ 5 / 3} + 15/2 x^{ 2 / 3} + C |

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15) a) f " (x) = 3x² + 2 | f '(x) = x³ + 2x - 1 | f(x) = 1/4 x^{ 4} + x² - x + 4 |

b) |

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16) v(t) = 2t - 3t² - 5, therefore s(t) = t² - t³ - 5t + 4

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