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MTH 121 — Summer — 2005 Essex County College — Division of Mathematics Test # 2^1 — Created June 10, 2005
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Show all work clearly and in order, and box your final answers. Justify your answers alge- braically whenever possible. You have at most 110 minutes^2 to take this 100 point exam. No cellular phones allowed.
- Use Newton’s method with the initial approximation x 1 = 1.000000 to find x 2 and x 3 , using six decimal places. 6 points x^3 − x^2 − 1 = 0
Solution: Let f (x) = x^3 − x^2 − 1 , where f ′^ (x) = 3x^2 − 2 x. Using Newton’s method as follows.
x 2 = x 1 − x
(^31) − x (^21) − 1 3 x^21 − 2 x 1
x 3 = x 2 −
x^32 − x^22 − 1 3 x^22 − 2 x 2 =
8 = 1.^625000
So, x 2 = 2. 000000 and x 3 = 1. 625000
- Find the critical numbers of the given function. 6 points
f (x) = 3x^4 + 4x^3 − 6 x^2
Solution: The first derivative is:
f ′^ (x) = 12x^3 + 12x^2 − 12 x = 12x
x^2 + x − 1
(^1) This document was prepared by Ron Bannon using LATEX 2ε. (^2) 8:30 a.m. until 10:20 a.m..
This is a polynomial and is differentiable everywhere. Now you need to determine where the derivative is zero.
12 x
x^2 + x − 1
= 0 ⇒ x = 0, x =
So, the critical numbers are:
− 1 −
- Find the point(s) on the ellipse 4x^2 + y^2 = 4 that is farthest from the point (1, 0). 12 points
In[6]:= Plot@Hx ^ 2 - 1 L ê Hx ^ 2 + 1 L, 8 x, - 5 , 5 <D
-4 -2 2 4
-0.
Out[6]= Ü Graphics Ü
In[14]:= Plot@8Sqrt@ 4 - 4 x ^ 2 D, - Sqrt@ 4 - 4 x ^ 2 D<, 8 x, - 1, 1 <, AspectRatio Ø Automatic, Epilog Ø 8 AbsolutePointSize@ 5 D, Poi
-1 -0.5 0.5 1
1
2
Figure 1: Graph of^ Out[14]=^ Ü^ Graphics f (x) on the interval [^ Ü − 5 , 5].
Solution: You’ll need to maximize the distance between the points (1, 0) and a point (x, y) on the ellipse. Clearly, the point on the ellipse in terms of x is
x, ±
4 − 4 x^2
There will be symmetry, so I’ll just analyze the upper half and use symmetry to get the lower half.
d = f (x) =
4 − 4 x^2 − 0
5 − 2 x − 3 x^2.
Here f ′^ (c) = 2π cos 2πc = 0 when c =
4 ,^ −^
4 ,^
4 ,^
- Given:
f (x) = x
x^2 + 1 f ′^ (x) = 4 x (x^2 + 1)^2
f ′′^ (x) =
1 − 3 x^2
(x^2 + 1)^3 In[6]:= Plot@Hx ^ 2 - 1 L ê Hx ^ 2 + 1 L, 8 x, - 5 , 5 <D
-4 -2 2 4
-0.
Out[6]= Ü Graphics Ü
Untitled-1 1
Figure 2: Graph of f (x) on the interval [− 5 , 5].
Answer the following questions.
(a) x-intercept(s): Answer: (− 1 , 0); (1, 0) 2 points
(b) y-intercept(s): Answer: (0, −1) 2 points
(c) vertical asymptote(s): Answer: none 2 points
(d) horizontal asymptote(s): Answer: y = 1 2 points
(e) domain: Answer: R 2 points
(f) range: Answer: [− 1 , 1) 2 points
(g) local maximum(s): Answer: none 2 points
(h) local minimum(s): Answer: (0, −1) 2 points
(i) global maximum(s): Answer: none 2 points
(j) global minimum(s): Answer: (0, −1) 2 points
(k) point(s) of inflection: Answer:
4 points
- Find the absolute minimum and absolute maximum values of f on the given interval. 7 points
f (x) = x^3 − 6 x^2 + 9x + 2 [− 1 , 4]
Solution: Use the Closed Interval Method. The derivative is: f ′^ (x) = 3x^2 − 12 x + 9 = 3 (x − 1) (x − 3). The critical numbers are 1 and 3.
f (−1) = − 14 f (1) = 6 f (3) = 2 f (4) = 6
So, the absolute minimum of f is f (−1) = −14 , and the absolute maximum of f is f (1) = f (4) = 6.
- Evaluate each of the following limits. 4 points (a)
xlim→∞
x^2 + 1 − x
Solution:
xlim→∞
x^2 + 1 − x
= (^) xlim→∞
x^2 + 1 − x 1 ·
√x^2 + 1 +^ x x^2 + 1 + x = (^) xlim→∞ √^1 x^2 + 1 + x
(d) Find the interval(s) where f is concave-down. 4 points
Solution: By sign analysis,
, 1 +^
- Find f if f ′^ (x) = x
x and f (1) = 2. 6 points
Solution:
f ′^ (x) = x
x = x
(^32)
f (x) = 2 x^
(^52)
5
+ C
f (1) = 2 5
+ C = 2, ⇒ C =^8
So,
f (x) =^2 x
2 √x + 8 5
