CALCULUS I TEST # 2 SOLUTIONS |

**Time limit:**

- 2 hours

**Instructions:**

- No graphing calculators

No notes or crib sheets allowed

Show all work. Write neatly and big enough to see!!

Numbers in parentheses ( ) are mark values.

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A/ Find y '. Do not simplify beyond basic algebra.

1) y' = (5/3)x^{ 2/3} sin 2x + 2x^{ 5/3} cos 2x |
4) y' = -14(2 cos 5x - 4 tan 3x)^{6} (5sin5x + 6 sec^{2} 3x) |

2) | 5) y = 3 sin^{2} 4x - 2 cos^{ 3} x^{2} |

3) |

(15)

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B/ 1)

which becomes *y = (16/11)x + 6/11*.

(4)

2) Evaluate the limits:

a) | b) |

(4)

3) *f (0) = 1*, *f (4) = 29, *and* f '(c) = 2c + 3*

Setting so *c = 2*.* *

(4)

4) Since A = o r^{ 2},* dA = 2 **o **r dr
*so the Area

(4)

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C/

1)

(4)

2)

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(5)

3) Draw the graph of and list the M_{2}I_{2}ACIDS features for

This function has a **hole in the graph at x = 3** since both numerator and denominator = 0.

We do the analysis on , then find the limit as x t 3 for the original function.

**Step 1: intercepts & asymptotes**

When x = 0, g(0) = –1/2 -- so (0, – 1/2) is *y*-intercept

**VA: x = **–** 2**, **HA: y = 0**

**Step 2: first derivative**

g '(x) < 0 ¼ x since the numerator is a negative constant, denominator is a perfect square.

g(x) is **decreasing ****¼**** x**, since g '(x) ! 0, **no max, no min**.

**Step 3: second derivative**

g "(x) ! 0, **no points of inflection.**

g "(x) < 0 when **x < **–** 2**, **concave down.**

g "(x) > 0 when **x > **–** 2**, **concave up.**

**Hole in the graph at x = 3**

, **hole at (3, 1/5)** -- only 1 hole -- so no symmetry.

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(10)

TOTAL (50)

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