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 sec2 3x) 2) 5) y = 3 sin2 4x - 2 cos 3 x2 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 decreases by 0.16
o inches² = 0.503 in 2.

(4)

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C/

1)

(4)

2)

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(5)

3) Draw the graph of and list the M2I2ACIDS 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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